Acids, Bases and Salts

All substances are acidic, neutral or basic (alkaline). How acidic or basic a substance is shown by its pH. There are several other ways by which we could find out whether a substance is acidic, neutral or basic.

pH Scale:

This is a scale that runs from 0 to 14. Substances with a pH below 7 are acidic. Substances with pH above 7 are basic. And those with pH 7 are neutral.

Indicators:

Indicators are substances that identify acidity or alkalinity of substances. They cannot be used in solid form.

Universal Indicator:

This is a substance that changes color when added to another substance depending on its pH. The indicator and the substance should be in aqueous form.

Litmus Paper or Solution:

This indicator is present in two colors: red and blue. We use blue litmus if we want test a substance for acidity. We use red litmus if we want to test a substance for alkalinity. Its results are:

Acids: Turns blue litmus paper/ solution red, Bases: Turns red litmus paper/ solution blue, Neutral: if it is used as paper the color doesn’t change. If it is used as solution it turns purple.

Note: use damp litmus paper if testing gases.

Phenolphthalein:

This is an indicator that is used to test for alkalinity because it is colorless if used with an acidic or neutral substance and it is pink if it is used with a basic substance.

Methyl Orange:

This indicator gives fire colors: Red with acids, yellow with neutrals and orange with bases.

 

Acids:

Acids are substances made of a hydrogen ion and non-metal ions.  They have the following properties:

They dissolve in water producing a hydrogen ion H+, They have a sour taste, Strong ones are corrosive, Their pH is less than 7.

All acids must be in aqueous form to be called an acid. For example Hydrochloric acid is hydrogen chloride gas dissolved in water. The most common acids are:

Hydrochloric acid HCl, Sulphuric Acid H2SO4, Nitric Acid HNO3, Cirtric Acid, Carbonic Acid H2CO3.

Strength of Acids:

One of the most important properties of acids is that it gives hydrogen ion H+ when dissolved in water. This is why the amount of H+ ions the acid can give when dissolved in water is what determines its strength. This is calledionization or dissociation. The more ionized the acid is the stronger it is, the lower its pH. The more H+ ions given when the acid is dissolved in water the more ionized the acid is.

Strong Acids:

Have pH’s: 0,1,2,3 They are fully ionized When dissolved in water, they give large amounts of H+ ionsExamples: Hydrochloric Acid Sulfuric Acid Nitric Acid

Weak Acids:

Have pH’s: 4,5,6 They are partially ionized When dissolved in water, they give small amounts of H+ ionsExamples: Ethanoic acid (CH3COOH) Citric Acid Carbonic Acid

Hydrochloric acid is a strong acid. When it is dissolved in water all HCl molecules are ionized into H+ and Cl- ions. It is fully ionized.

Ethanoic acid has the formula CH3COOH. It is a weak acid. When it is dissolved in water, only some of the CH3COOH molecules are ionized into CH3COO-  and H+ ions. It is partially ionized.

Note: Acids with pH 3 or 4 can be considered moderate in strength.

Solutions of strong acids are better conductors of electricity than solutions of weak acids. This is because they contain much more free mobile ions to carry the charge.

Concentrated acids are not necessarily strong. The concentration of an acid only means the amount of molecules of the acid dissolved in water. Concentrated acids have a large amount of acid molecules dissolved in water. Dilute acids have a small amount of acid molecules dissolved in water. Concentration is not related to strength of the acids. Strong acids are still strong even if they are diluted. And weak acids are still weak even if they are concentrated.

 

Bases:

Bases are substances made of hydroxide OH- ions and a metal. Bases can be made of:

Metal hydroxide (metal ion & OH- ion) Metal oxides Metal carbonates (metal ion & CO3

2-) Metal hydrogen carbonate (Bicarbonate) Ammonium hydroxide (NH4OH) Ammonium Carbonate ((NH4)2CO3)

Properties of bases:

Bitter taste Soapy feel Have pH’s above 7 Strong ones are corrosive

Some bases are water soluble and some bases are water insoluble. Water soluble bases are also called alkalis.

Like acids, alkalis' strength is determined by its ability to be ionized into metal and hydroxide OH-  ions. Completely ionized alkalis are the strongest and partially ionized alkalis are the weakest. Ammonium hydroxide is one of the strongest alkalis while weak alkalis include the hydroxides of sodium, potassium and magnesium.

 

Types of Oxides:

Basic Oxides

They are metal oxides They react with acids forming a salt and water They are solids They are insoluble in water except group 1

Amphoteric Oxides

These are oxides of Aluminum, Zinc & Lead They act as an acid when reacting with an alkali & vice versa Their element’s

Acidic Oxides

They are all non-metal oxides except non-metal monoxides They are gases They react with an alkali to form salt and

metal oxides. They react with an acid forming salt and water Examples: Na2O, CaO and CuO

hydroxides are amphoteric too They produce salt and water when reacting with an acid or an alkali.

water Note: metal monoxides are neutral oxides Examples: CO2, NO2, SO2 (acidic oxides) & CO, NO,H2O (neutral oxides)

 

Salts:

A salt is a neutral ionic compound. Salts are one of the products of a reaction between an acid and a base. Salts are formed in reactions I n which the H+ ion from the acid is replaced by any other metal ion. Some salts are soluble in water and some are insoluble.

Soluble Salts:

All Nitrates All halides EXCEPT AgCl and PbCl2 All sulfates EXCEPT CaSO4, BASO4, PbSO4 All group 1 metals salts All ammonium salts

Insoluble Salts:

Silver and lead chlorides (AgCl & PbCl2) Calcium, barium and lead sulphates (CaSO4, BASO4, PbSO4) All carbonates EXCEPT group 1 metals and ammonium carbonates

 

Preparing Soluble Salts:

Displacement Method (Excess Metal Method):

Metal + Acid → Salt + Hydrogen

Note: this type of method is suitable to for making salts of moderately reactive metals because highly reactive metals like K, Na and Ca will cause an explosion. This method is used with the MAZIT (Magnesium, Aluminum, Zinc, Iron and Tin) metals only.

Example: set up an experiment to obtain magnesium chloride salt.

Mg + 2HCl → MgCl2 + H2

1. Add 100 cm3 of dilute hydrochloric acid to a beaker2. Add excess mass of powdered magnesium3. When the reaction is done, filter the mixture to get rid of excess magnesium (residue)4. The filtrate is magnesium chloride solution5. To obtain magnesium chloride powder, evaporate the solution till dryness6. To obtain magnesium chloride crystals, heat the solution while continuously dipping a glass rod in the

solution7. When you observe crystals starting to form on the glass rod, turn heat off and leave the mixture to cool down

slowly8. When the crystals are obtained, dry them between two filter papers

Observations of this type of reactions:

Bubbles of colorless gas evolve (hydrogen). To test approach a lighted splint if hydrogen is present it makes a pop sound

The temperature rises (exothermic reaction) The metal disappears

You know the reaction is over when:

No more gas evolves No more magnesium can dissolve The temperature stops rising The solution becomes neutral

Proton Donor and Acceptor Theory:

When an acid and a base react, water is formed. The acid gives away an H+ ion and the base accepts it to form water by bonding it with the OH- ion. A hydrogen ion is also called a proton this is why an acid can be called Proton Donor and a base can be called Proton Acceptor.

 

Neutralization Method:

Acis + Base → Salt + Water

Note: This method is used to make salts of metals below hydrogen in the reactivity series. If the base is a metal oxide or metal hydroxide, the products will be salt and water only. If the base is a metal carbonate, the products will be salt, water and carbon dioxide.

Type 1:

Acid + Metal Oxide → Salt + Water

To obtain copper sulfate salt given copper oxide and sulfuric acid:

CuO + H2SO4 → CuSO4 + H2O

Add 100 cm3 of sulfuric acid to a beaker Add excess mass of Copper oxide When the reaction is over, filter the excess copper oxide off The filtrate is a copper sulfate solution, to obtain copper sulfate powder evaporate the solution till dryness To obtain copper sulfate crystals, heat the solution white continuously dipping a glass rod in it When you observe crystals starting to form on the glass rod, turn heat of and leave the mixture to cool down

slowly When you obtain the crystals dry them between two filter papers

Observations of this reaction:

The amount of copper oxide decreases The solution changes color from colorless to blue The temperature rises You know the reaction is over when No more copper oxide dissolves The temperature stops rising The solution become neutral

Type 2:

Acid + Metal Hydroxide → Salt + Water

to obtain sodium chloride crystals given sodium hydroxide and hydrochloric acid:

HCl + NaOH → NaCl + H2O

Add 100 cm3of dilute hydrochloric acid to a beaker Add excess mass of sodium hydroxide When the reaction is over, filter the excess sodium hydroxide off The filtrate is sodium chloride solution, to obtain sodium chloride powder, evaporate the solution till dryness To obtain sodium chloride crystals, hear the solution while continuously dipping a glass rod in it When crystals start to form on the glass rod, turn heat off and leave the mixture to cool down slowly When the crystals are obtained, dry them between two filter papers

Observations:

Sodium hydroxide starts disappearing Temperature rises

You know the reaction is over when:

The temperature stops rising No more sodium hydroxide can dissolve The pH of the solution becomes neutral

Type 3:

Acid + Metal Carbonate → Salt + Water + Carbon Dioxide

To obtain copper sulfate salt given copper carbonate and sulfuric acid:

CuCO3 + H2SO4 → CuSO4 + H2O + CO2

Add 100 cm3 of dilute sulfuric acid to a beaker Add excess mass of copper carbonate When the reaction is over, filter excess copper carbonate off The filtrate is a copper sulfate solution, to obtain copper sulfate powder evaporate the solution till dryness To obtain copper sulfate crystals, heat the solution white continuously dipping a glass rod in it When you observe crystals starting to form on the glass rod, turn heat of and leave the mixture to cool down

slowly When you obtain the crystals dry them between two filter papers

Observations:

Bubbles of colorless gas (carbon dioxide) evolve, test by approaching lighted splint, if the CO2  is present the flame will be put off

Green Copper carbonate starts to disappear The temperature rises The solution turns blue

You know the reaction is finished when:

No more bubbles are evolving The temperature stops rising No more copper carbonate can dissolve The pH of the solution becomes neutral

Titration Method:

This is a method to make a neutralization reaction between a base and an acid producing a salt without any excess. In this method, the experiment is preformed twice, the first time is to find the amounts of reactants to use, and the second experiment is the actual one.

1st Experiment:

Add 50 cm3 of sodium hydroxide using a pipette to be accurate to flask

Add 5 drops of phenolphthalein indicator to the sodium hydroxide. The solution turns pink indicating presence of a base

Fill a burette to zero mark with hydrochloric acid Add drops of the acid to conical flask The pink color of the solution becomes lighter When the solution turns colorless, stop adding the acid (End point:

is the point at which every base molecule is neutralized by an acid molecule)

Record the amount of hydrochloric acid used and repeat the experiment without using the indicator

After the 2nd experiment, you will have a sodium chloride solution. Evaporate it till dryness to obtain powdered sodium chloride or crystalize it to obtain sodium chloride crystals

 

Preparing Insoluble Salts:

Precipitation Method:

A precipitation reaction is a reaction between two soluble salts. The products of a precipitation reaction are two other salts, one of them is soluble and one is insoluble (precipitate).

Example: To obtain barium sulfate salt given barium chloride and sodium sulfate:

BaCl2 + Na2SO4 → BaSO4 + 2NaClIonic Equation: Ba2+ + SO4

2- → BaSO4

Add the two salt solutions in a beaker When the reaction is over, filter and take the residue Wash the residue with distilled water and dry it in the oven

Observations:

Temperature increases An insoluble solid precipitate (Barium sulfate) forms

You know the reaction is over when:

The temperature stops rising No more precipitate is being formed

 

Controlling Soil pH:

If the pH of the soil goes below or above 7, it has to be neutralized using an acid or a base. If the pH of the soil goes below 7, calcium carbonate (lime stone) is used to neutralize it. The pH of the soil can be measured by taking a sample from the soil, crushing it, dissolving in water then measuring the pH of the solution.

Colors of Salts:

Salt Formula Solid In Solution

Hydrated copper sulfate CuSO4.5H2O Blue crystals Blue

Anhydrous copper sulfate CuSO4 White powder Blue

Copper nitrate Cu(NO3)2 Blue crystals Blue

Copper chloride CuCl2 Green Green

Copper carbonate CuCO3 Green Insoluble

Copper oxide CuO Black Insoluble

Iron(II) salts E.g.: FeSO4, Fe(NO3)2 Pale green crystals Pale green

Iron(III) salts E.g.: Fe(NO3)3 Reddish brown Reddish brown

 

Tests for Gases:

Gas Formula Tests

Ammonia NH3 Turns damp red litmus paper blue

Carbon dioxide CO2 Turns limewater milky

Oxygen O2 Relights a glowing splint

Hydrogen H2 ‘Pops’ with a lighted splint

Chlorine Cl2 Bleaches damp litmus paper

Nitrogen dioxide NO2 Turns damp blue litmus paper red

Sulfur dioxide SO2 Turns acidified aqueous potassium dichromate(VI) from orange to green

 

Tests for Anions:

Anion Test Result

Carbonate (CO32-) Add dilute acid Effervescence,

carbon dioxide produced

Chloride (Cl-)(in solution)

Acidify with dilute nitric acid, then addaqueous silver nitrate White ppt.

Iodide (I-)(in solution)

Acidify with dilute nitric acid, then addaqueous silver nitrate Yellow ppt.

Nitrate (NO3-)(in solution)

Add aqueous sodium hydroxide, thenaluminium foil; warm carefully Ammonia produced

Sulfate (SO42-) Acidify, then add aqueous barium nitrate White ppt.

 

Tests for aqueous cations:

Cation Effect of aqueous sodium hydroxide Effect of aqueous ammonia

Aluminium (Al3+) White ppt., soluble in excess giving acolourless solution White ppt., insoluble in excess

Ammonium (NH4+) Ammonia produced on warming –

Calcium (Ca2+) White ppt., insoluble in excess No ppt. or very slight white ppt.

Copper (Cu2+) Light blue ppt., insoluble in excess Light blue ppt., soluble in excess,giving a dark blue solution

Iron(II) (Fe2+) Green ppt., insoluble in excess Green ppt., insoluble in excess

Iron(III) (Fe3+) Red-brown ppt., insoluble in excess Red-brown ppt., insoluble in excess

Zinc (Zn2+) White ppt., soluble in excess,giving a colourless solution

White ppt., soluble in excess,giving a colourless solution

Acid-Base Titrations

Acid-Base titrations are usually used to find the the amount of a known acidic or basic substance through acid base reactions. The analyte (titrand) is the solution with an unknown molarity. The reagent (titrant) is the solution with a known molarity that will react with the analyte.

ProcedureThe analyte is prepared by dissolving the substance being studied into a solution. The solution is usually placed in a flask for titration. A small amount of indicator is then added into the flask along with the analyte. The reagent is usually placed in a burette and slowly added to the analyte and indicator mixture. The amount of reagent used is recorded when the indicator causes a change in the color of the solution.Some titrations requires the solution to be boiled due to the CO2 created from the acid-base reaction. The CO2 forms carboinic acid (H2CO3) when dissolved in water. The carbonic acid then acts as a buffer, reducing the accuracy of data. After boiling most of the CO2will be removed from the solution allowing the solution to be titrated to a more accurate endpoint. The endpoint is the point where all of the analyte has be reacted with the reagent.

Choosing an Indicator A useful indicator has a strong color that changes quickly near its pKa. These traits are desirable so only a small amount of an indicator is needed. If a large amount of indicator is used, the indicator will effect the final pH, lowering the accuracy of the experiment. The indicator should also have a pKa value near the pH of the titration's endpoint. For example a analyte that is a weak base would require an indicator with a pKa less than 7. Choosing an indicator with a pKa near the endpoint's pH will also reduce error because the color change occurs sharply during the endpoint where the pH spikes, giving a more precise endpoint.

Figure 1: A Basic Titration Curve

Notice that this reaction is between a weak acid and a strong base so phenolphthalein with a pKa of 9.1 would be a better choice than methyl orange with a pKa of 3.8. If in this reaction we were to use methyl orange as the indicator color changes would occur all throughout the region highlighted in pink. The data obtained would be hard to determine due to the large range of color change, and inaccurate as the color change does not even lie with the endpoint region. Phenolphthalein on the other hand changes color rapidly near the endpoint allowing for more accurate data to be gathered.

CalculationsMultiply the volume of reagent added to get to the endpoint, with the molarity of the reagent to find the moles of reagent used. With the balanced equation of the acid-base reaction in question to find the moles of unknown substance. Then the original molarity can be calculated by dividing through with the initial volume.

For example an unknown molarity of HCl acts as the analyte. 50mL of it is placed into a flask and a 0.1M solution of NaOH will be the reagent. The endpoint's pH is 7 so litmus, with a pKa of 6.5 is chosen. The color of the solution changes when 10mL of 0.1M NaOH is added.

The balanced chemical equation related to isHCl(aq)+NaOH(aq)→H2O(l)+Na++Cl−

Or just the net ionic equationH++OH−→H2O(l)

The following equation can then be derived

        

        X = 0.0010 mol of HCl

The molarity is now easily solved for

        0.0010 mol HCl / 0.050 L = 0.020M HCl

Buffers

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.

What is a buffer composed of?To effectively maintain a pH range, a buffer must consist of a weak conjugate acid-base pair, meaning either a. a weak acid and its conjugate base, or b. a weak base and its conjugate acid. The use of one or the other will simply depend upon the desired  pH when preparing the buffer. For example, the following could function as buffers when together in solution:

Acetic acid (weak organic acid w/ formula CH3COOH) and a salt containing its conjugate base, the acetate anion (CH3COO-), such as sodium acetate (CH3COONa)

Pyridine (weak base w/ formula C5H5N) and a salt containing its conjugate acid, the pyridinium cation (C 5H5NH+), such as Pyridinium Chloride.

Ammonia (weak base w/ formula NH3) and a salt containing its conjugate acid, the ammonium cation, such as Ammonium Hydroxide (NH4OH)

How does a buffer work?A buffer is able to resist pH change because the two components (conjugate acid and conjugate base) are both present in appreciable amounts at equilibrium and are able to neutralize small amounts of other acids and bases (in the form of H 3O+ and OH-) when the are added to the solution.In order to clarify this effect, we can consider the simple example of a Hydrofluoric Acid (HF) and Sodium Fluoride (NaF) buffer. Hydrofluoric acid is a weak acid due to the strong attraction between the relatively small F - ion and solvated protons (H3O+), which does not allow it to dissociate completely in water. Therefore, if we obtain HF in an aqueous solution, we establish the following equilibrium with only slight dissociation (Ka(HF) = 6.6x10-4, strongly favors reactants):

HF(aq)+H2O(l)⇌F−(aq)+H3O+(aq)

We can then add and dissolve sodium fluoride into the solution and mix the two until we reach the desired volume and pH at which we want to buffer.

When Sodium Fluoride dissolves in water, the reaction goes to completion, thus we obtain:

NaF(aq)+H2O(l)→Na+(aq)+F−(aq)

Since Na+ is the conjugate of a strong base, it will have no effect on the pH or reactivity of the buffer. The addition of NaF to the solution will, however, increase the concentration of F- in the buffer solution, and, consequently, by Le Chatlier's principle, lead to slightly less dissociation of the HF in the previous equilibrium, as well.  The presence of significant amounts of both the conjugate acid, HF, and the conjugate base, F-, allows the solution to function as a buffer.This buffering action can be seen in the titration curve of a buffer solution.

As we can see, over the working range of the buffer. pH changes very little with the addition of acid or base. Once the buffering capacity is exceeded the rate of pH change quickly jumps. This occurs because the conjugate acid or base has been depleted throughneutralization. This principle implies that a larger amount of conjugate acid or base will have a greater buffering capacity.If acid is added:

F−(aq)+H3O+(aq)⇌HF(aq)+H2O(l)

In this reaction, the conjugate base, F-, will neutralize the added acid, H3O+, and this reaction goes to completion, because the reaction of F- with H3O+ has an equilibrium constant much greater than one.  (In fact, the equilibrium constant the reaction as written is just the inverse of the Ka for HF: 1/Ka(HF) = 1/(6.6x10-4) = 1.5x10+3.)  So long as there is more F- than H3O+, almost all of the H3O+ will be consumed and the equilibrium will shift to the right, slightly increasing the concentration of HF and slightly decreasing the concentration of F-, but resulting in hardly any change in the amount of H3O+ present once equilibrium is re-established. If base is added:

HF(aq)+OH−(aq)⇌F−(aq)+H2O(l)

In this reaction, the conjugate acid, HF, will neutralize added amounts of base, OH -, and the equilibrium   will again shift to the right, slightly increasing the concentration of F- in the solution and decreasing the amount of HF slightly. Again, since most of the OH - is neutralized, little pH change will occur.

These two reactions can continue to alternate back and forth with little pH change.

Selecting proper components for desired pHBuffers function best when the pKa of the conjugate weak acid used is close to the desired working range of the buffer. This turns out to be the case when the concentrations of the conjugate acid and conjugate base are approximately equal (within about a factor of 10). For example, we know the Ka for hydroflouric acid is 6.6 x 10-4 so its pKa= -log(6.6 x 10-4) = 3.18. So, a hydrofluoric acid buffer would work best in a buffer range of around pH = 3.18.For the weak base ammonia (NH3), the value of Kb is 1.8x10-5, implying that the Ka for the dissociation of its conjugate acid, NH4

+, is Kw/Kb=10-14/1.8x10-5 = 5.6x10-10.  Thus, the pKa for NH4

+ = 9.25, so buffers using NH4+/NH3 will work best around a pH of 9.25. (It's always

the pKa of the conjugate acid that determines the approximate pH for a buffer system, though this is dependent on the pK b of the conjugate base, obviously.)When the desired pH of a buffer solution is near the pKa of the conjugate acid being used (i.e., when the amounts of conjugate acid and conjugate base in solution are within about a factor of 10 of each other), the Henderson-Hasselbalch equation can be applied as a simple approximation of the solution pH, as we will see in the next section.

Example: HF Buffer

In this example we will continue to use the hydrofluoric acid buffer. We will discuss the process for preparing a buffer of HF at a  pH of 3.0. We can use the Henderson-Hasselbalch equation to calculate the necessary ratio of F- and HF.

pH=pKa+log[Base][Acid]

3.0=3.18+log[Base][Acid]

log[Base][Acid]=−0.18

[Base][Acid]=10−0.18

[Base][Acid]=0.66

This is simply the ratio of the concentrations of conjugate base and conjugate acid we will need in our solution. However, what if we have 100 ml of 1 M HF and we want to prepare a buffer using NaF? How much Sodium Fluoride would we need to add in order to create a buffer at said pH (3.0)?

We know from our Henderson-Hasselbalch calculation that the ratio of our base/acid should be equal to 0.66.  From a table of molar masses, such as a periodic table, we can calculate the molar mass of NaF to be equal to 41.99 g/mol. HF is a weak acid with a Ka = 6.6 x 10-4 and the concentration of HF is given above as 1 M. Using this information, we can calculate the amount of F- we need to add.The dissociation reaction is:

HF(aq)+H2O(l)⇌F−(aq)+H3O+(aq)

We could use ICE tables to calculate the concentration of F- from HF dissociation, but, since Ka is so small, we can approximate that virtually all of the HF will remain undissociated, so the amount of F - in the solution from HF dissociation will be negligible.  Thus, the [HF] is about 1 M and the [F -] is close to 0.  This will be especially true once we have added more F -, the addition of which will even further suppress the dissociation of HF. We want the ratio of Base/Acid to be 0.66, so we will need [Base]/1M = 0.66.  Thus, [F-] should be about 0.66 M.  For 100 mL of solution, then, we will want to add 0.066 moles (0.1 L x 0.66 M) of F -.  Since we are adding NaF as our source of F-, and since NaF completely dissociates in water, we need 0.066 moles of NaF.  Thus, 0.066 moles x 41.99 g/mol = 2.767 g.Note that, since the conjugate acid and the conjugate base are both mixed into the same volume of solution in the buffer, the ratio of "Base/Acid" is the same whether we use a ratio of the "concentration of base over concentration of acid," OR a ratio of "moles of base over moles of acid."  The pH of the solution does not, it turns out, depend on the volume!  (This is only true so long as the solution does not get so dilute that the autoionization of water becomes an important source of H + or OH-.  Such dilute solutions are rarely used as buffers, however.)

Adding Strong Acids or Bases to Buffer SolutionsNow that we have this nice F-/HF buffer, let's see what happens when we add strong acid or base to it. Recall that the amount of F- in the solution is 0.66M x 0.1 L = 0.066 moles and the amount of HF is 1.0 M x 0.1L = 0.10 moles. Let's double check the pH using theHenderson-Hasselbalch Approximation, but using moles instead of concentrations:

pH = pKa + log(Base/Acid) = 3.18 + log(0.066 moles F-/0.10 moles HF) = 3.00Good.  Now let's see what happens when we add a small amount of strong acid, such as HCl. When we put HCl into water, it completely dissociates into H3O+ and Cl-.  The Cl- is the conjugate base of a strong acid so is inert and doesn't affect pH, and we can just ignore it. However, the H3O+ can affect pH and it can also react with our buffer components. In fact, we already discussed what happens. The equation is:

F−(aq)+H3O+(aq)⇌HF(aq)+H2O(l)

For every mole of H3O+ added, an equivalent amount of the conjugate base (in this case, F-) will also react, and the equilibrium constant for the reaction is large, so the reaction will continue until one or the other is essentially used up. If the F - is used up before reacting away all of the H3O+, then the remaining H3O+ will affect the pH directly. In this case, the capacity of the buffer will have been exceeded - a situation one tries to avoid. However, for our example, let's say that the amount of added H 3O+ is smaller than the amount of F- present, so our buffer capacity is NOT exceeded.  For the purposes of this example, we'll let the added H3O+ be equal to 0.01 moles (from 0.01 moles of HCl). Now, if we add 0.01 moles of HCl to 100 mL of pure water, we would expect the pH of the resulting solution to be 1.00 (0.01 moles/0.10 L = 0.1 M; pH = -log(0.1) = 1.0). However, we are adding the H3O+ to a solution that has F- in it, so the H3O+ will all be consumed by reaction with F-. In the process, the 0.066 moles of F- is reduced:

0.066 initial moles F- - 0.010 moles reacted with H3O+ = 0.056 moles F- remainingAlso during this process, more HF is formed by the reaction:

0.10 initial moles HF + 0.010 moles from reaction of F- with H3O+ = 0.11 moles HF after reactionPlugging these new values into Henderson-Hasselbalch gives:

pH = pKa + log (base/acid) = 3.18 + log (0.056 moles F-/0.11 moles HF) = 2.89Thus, our buffer did what it should - it resisted the change in pH, dropping only from 3.00 to 2.89 with the addition of 0.01 moles of strong acid.

Acid and Base IndicatorsThe most common method to get an idea about the pH of solution is to use an acid base indicator. An indicator is a large organic molecule that works somewhat like a " color dye". Whereas most dyes do not change color with the amount of acid or base present, there are many molecules, known as acid - base indicators , which do respond to a change in the hydrogen ion concentration. Most of the indicators are themselves weak acids.

IndicatorsThe most common indicator is found on "litmus" paper. It is red below pH 4.5 and blue above pH 8.2.

 Color  Blue Litmus Red Litmus

 Acid  turns red stays same

 Base  stays same turns blue

Other commercial pH papers are able to give colors for every main pH unit. Universal Indicator, which is a solution of a mixture of indicators is able to also provide a full range of colors for the pH scale.

A variety of indicators change color at various pH levels. A properly selected acid-base indicator can be used to visually "indicate" the approximate pH of a sample. An indicator is usually some weak organic acid or base dye that changes colors at definite pH values. The weak acid form (HIn) will have one color and the weak acid negative ion (In-) will have a different color. The weak acid equilibrium is:

HIn → H+ + In-

For phenolphthalein: pH 8.2 = colorless; pH 10 = red For bromophenol blue: pH 3 = yellow; pH 4.6 = blue

See the graphic below for colors and pH ranges.

Magic Pitcher DemonstrationPhenolphthalein is an indicator of acids (colorless) and bases (pink). Sodium hydroxide is a base, and it was in the pitcher at the beginning, so when added to the phenolphthalein in beakers 2 and 4, it turned pink (top half of the graphic).

ReactionEquilibrium: HIn → H+ + In-

colorless              red

The equilibrium shifts right, HIn decreases, and In - increases. As the pH increase between 8.2 to 10.0 the color becomes red because of the equilibrium shifts to form mostly In- ions.

The third beaker has only the NaOH but no phenolphthalein, so it remained colorless. The first beaker contain acetic acid and is skipped over at first.

After pouring beakers 2, 3, 4 back into the pitcher it give a pink solution. Bottom half of the graphic: When the pitcher is then poured back into beakers 2, 3, 4 it is a pink solution. In the first beaker, a strange thing happens in that the pink solution coming out of the pitcher now changes to colorless. This happens

because the first beaker contains some vinegar or acetic acid which neutralizes the NaOH, and changes the solution from basic to acidic. Under acidic conditions, the phenolphthalein indicator is colorless.

Neutralization reaction:

HC2H3O2 + NaOH → Na(C2H3O2) + HOH

Explain the color indicator changeUse equilibrium principles to explain the color change for phenolphthalein at the end of the demonstration.

Solution

The simplified reaction is: H+ + OH- → HOH

As OH- ions are added, they are consumed by the excess of acid already in the beaker as expressed in the above equation. The hydroxide ions keep decreasing and the hydrogen ions increase, pH decreases.

See lower equation: The indicator equilibrium shifts left, In- ions decrease. Below pH 8.2 the indicator is colorless. As H+ ions are further increased and pH decreases to pH 4-5, the indicator equilibrium is effected and changes to the colorless HIn form.

Equilibrium: HIn → H+ + In-

colorless red

 

Molecular Basis for the Indicator Color ChangeColor changes in molecules can be caused by changes in electron confinement. More confinement makes the light absorbed more blue, and less makes it more red.

How are electrons confined in phenolphthalein? There are three benzene rings in the molecule. Every atom involved in a double bond has a p orbital which can overlap side-to-side with similar atoms next to it. The overlap creates a 'pi bond' which allows the electrons in the p orbital to be found on either bonded atom. These electrons can spread like a cloud over any region of the molecule that is flat and has alternating double and single bonds. Each of the benzene rings is such a system.

See the far left graphic - The carbon atom at the center (adjacent to the yellow circled red oxygen atom) doesn't have a p-orbital available for pi-bonding, and it confines the pi electrons to the rings. The molecule absorbs in the ultraviolet, and this form of phenolphthalein is colorless.

In basic solution, the molecule loses one hydrogen ion. Almost instantly, the five-sided ring in the center opens and the electronic structure around the center carbon changes (yellow circled atoms) to a double bond which now does contain pi electrons. The pi electrons are no longer confined separately to the three benzene rings, but because of the change in geometry around the yellow circled atoms, the whole molecule is now flat and electrons are free to move within the entire molecule. The result of all of these changes is the change in color to pink. Chime: Phenolphthalein

Many other indicators behave on the molecular level in a similar fashion (the details may be different) but the result is a change in electronic structure along with the removal of a hydrogen ion from the molecule. Plant pigments in flowers and leaves also behave in this fashion.

Brønsted Concept of Acids and BasesTable of Contents

1. 1. Brønsted-Lowery Definition2. 2. Acids are Proton Donors and Bases are Proton Acceptors3. 3. Questions4. 4. Outside Links5. 5. Sources6. 6. Contributors

In 1923, chemists Johannes Brønsted and Martin Lowry independently developed definitions of acids and bases based on compounds abilities to either donate or accept protons (H+ ions). Here, acids are defined as being able to donate protons in the form of hydrogen ions; whereas bases are defined as being able to accept protons. This took the Arrhenius definition one step further as water is no longer required to be present in the solution for acid and base reactions to occur.

Brønsted-Lowery DefinitionIn 1923, J.N. Brønsted and T.M. Lowry independently developed the theory of proton donors and proton acceptors in acid-base reactions, coincidentally in the same region and during the same year. The Arrhenius theory where acids and bases are defined by whether the molecule contains hydrogen and hydroxide ion is too limiting. The main effect of the Brønsted-Lowry definition is to identify the proton (H+) transfer occurring in the acid-base reaction. This is best illustrated in the following equation:

 HA + Z ↔ A- + HZ+

 Acid  Base  

 Donates hydrogen ions  Accepts hydrogen ions.  

 HCl+  HOH → H3O+ + Cl-

  HOH+   NH3→  NH4+ + OH-

The determination of a substance as a Brønsted-Lowery acid or base can only be done by observing the reaction. In the case of the HOH it is a base in the first case and an acid in the second case.

To determine whether a substance is an acid or a base, count the hydrogens on each substance before and after the reaction. If the number of hydrogens has decreased that substance is the acid (donates hydrogen ions). If the number of hydrogens has increased that substance is the base (accepts hydrogen ions). These definitions are normally applied to the reactants on the left. If the reaction is viewed in reverse a new acid and base can be identified. The substances on the right side of the equation are called conjugate acid and conjugate base compared to those on the left. Also note that the original acid turns in the conjugate base after the reaction is over.

Acids are Proton Donors and Bases are Proton AcceptorsSo what does this mean? For a reaction to be in equilibrium a transfer of electrons needs to occur. The acid will give an electron away and the base will receive the electron. Acids and Bases that work together in this fashion are called a conjugate pair made up ofconjugate acids and conjugate bases.

HA+Z⇌A−+HZ+

A stands for an Acidic compound and Z stands for a Basic compound

A Donates H to form HZ+. Z Accepts H from A which forms HZ+

A- becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibrium HZ+ becomes a conjugate acid of Z and in the reverse reaction it donates a H to A- recreating Z in order to remain in equilibrium

Brønsted Concept of Acids and BasesTable of Contents

1. 1. Brønsted-Lowery Definition2. 2. Acids are Proton Donors and Bases are Proton Acceptors3. 3. Questions4. 4. Outside Links5. 5. Sources6. 6. Contributors

In 1923, chemists Johannes Brønsted and Martin Lowry independently developed definitions of acids and bases based on compounds abilities to either donate or accept protons (H+ ions). Here, acids are defined as being able to donate protons in the form of hydrogen ions; whereas bases are defined as being able to accept protons. This took the Arrhenius definition one step further as water is no longer required to be present in the solution for acid and base reactions to occur.

Brønsted-Lowery DefinitionIn 1923, J.N. Brønsted and T.M. Lowry independently developed the theory of proton donors and proton acceptors in acid-base reactions, coincidentally in the same region and during the same year. The Arrhenius theory where acids and bases are defined by whether the molecule contains hydrogen and hydroxide ion is too limiting. The main effect of the Brønsted-Lowry definition is to identify the proton (H+) transfer occurring in the acid-base reaction. This is best illustrated in the following equation:

 HA + Z ↔ A- + HZ+

 Acid  Base  

 Donates hydrogen ions  Accepts hydrogen ions.  

 HCl+  HOH → H3O+ + Cl-

  HOH+   NH3→  NH4+ + OH-

The determination of a substance as a Brønsted-Lowery acid or base can only be done by observing the reaction. In the case of the HOH it is a base in the first case and an acid in the second case.

To determine whether a substance is an acid or a base, count the hydrogens on each substance before and after the reaction. If the number of hydrogens has decreased that substance is the acid (donates hydrogen ions). If the number of hydrogens has increased that substance is the base (accepts hydrogen ions). These definitions are normally applied to the reactants on the left. If the reaction is viewed in reverse a new acid and base can be identified. The substances on the right side of the equation are called conjugate acid and conjugate base compared to those on the left. Also note that the original acid turns in the conjugate base after the reaction is over.

Acids are Proton Donors and Bases are Proton AcceptorsSo what does this mean? For a reaction to be in equilibrium a transfer of electrons needs to occur. The acid will give an electron away and the base will receive the electron. Acids and Bases that work together in this fashion are called a conjugate pair made up ofconjugate acids and conjugate bases.

HA+Z⇌A−+HZ+

A stands for an Acidic compound and Z stands for a Basic compound

A Donates H to form HZ+. Z Accepts H from A which forms HZ+

A- becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibrium HZ+ becomes a conjugate acid of Z and in the reverse reaction it donates a H to A- recreating Z in order to remain in equilibrium

Lewis Concept of Acids and BasesAcids and bases are an important part of chemistry. One of the most applicable theories is the Lewis acid/base motif that extends the definition of an acid and base beyond H+ and OH- ions as described by Br ø nsted-Lowry   acids and bases .

IntroductionThe Brønsted acid-base theory has been used throughout the history of acid and base chemistry.  However, this theory is very restrictive and focuses primarily on acids and bases acting as proton donors and acceptors. Sometimes conditions arise where the theory doesn't necessarily fit, such as in solids and gases. In 1923, G.N. Lewis from UC Berkeley proposed an alternate theory to describe acids and bases. His theory gave a generalized explanation of acids and bases based on structure and bonding. Through the use of the Lewis definition of acids and bases, chemists are now able to predict a wider variety of acid-base reactions. Lewis' theory used electrons instead of proton transfer and specifically stated that an acid is a species that accepts an electron pair while a base donates an electron pair.

Figure 1 Above: A Lewis Base (B) donates it electrons to a Lewis Acid (A) resulting in a coordinate covalently bonded compound, also known as an adduct.

The reaction of a Lewis acid and a Lewis base will produce a coordinate covalent bond, as shown in Figure 1 above. A coordinate covalent bond is just a type of covalent bond in which one reactant gives it electron pair to another reactant. In this case the lewis base donates its electrons to the lewis acid. When they do react this way the resulting product is called an addition compound, or more commonly an adduct.

Lewis Acid: a species that accepts an electron pair (i.e., an electrophile) and will have vacant orbitals Lewis Base: a species that donates an electron pair (i.e., a nucleophile) and will have lone-pair electrons

Lewis AcidsLewis acids accept an electron pair. Lewis Acids are Electrophilic   meaning that they are electron attracting. When bonding with a base the acid uses its lowest unoccupied molecular orbital or LUMO (see Fig. 2).

Various species can act as Lewis acids. All cations are Lewis acids since they are able to accept electrons. (e.g., Cu2+, Fe2+, Fe3+) An atom, ion, or molecule with an incomplete octet of electrons can act as an Lewis acid (e.g., BF3, AlF3). Molecules where the central atom can have more than 8 valence shell electrons can be electron acceptors, and thus are classified as

Lewis acids (e.g., SiBr4, SiF4).

Molecules that have multiple bonds between two atoms of different electronegativities (e.g., CO2, SO2)

Lewis BasesLewis Bases donate an electron pair. Lewis Bases are Nucleophilic meaning that they “attack” a positive charge with their lone pair. They utilize the highest occupied molecular orbital or HOMO (see Fig. 2). An atom, ion, or molecule with a lone-pair of electrons can thus be a Lewis base. Each of the following anions can "give up" their electrons to an acid.  

Example: OH-, CN-, CH3COO-, :NH3, H2O:, CO:

HOMO & LUMOLewis base's HOMO (highest occupied molecular orbital) interacts with the Lewis acid's LUMO (lowest unoccupied molecular orbital) to create bonded molecular orbitals. Both Lewis Acids and Bases contain HOMO and LUMOs but only the HOMO is considered for Bases and only the LUMO is considered for Acids. (see fig. 2 below)

 

Figure 2 Above: Lewis Acids have vacant orbitals so they are in a lower energy level. While lewis bases have lone pair electrons to share and thus occupy a higher energy level.

Complex Ion / Coordination CompoundsComplex ions are polyatomic ions, which are formed from a central metal ion that has other smaller ions joined around it. While Brønsted theory can't explain this reaction Lewis acid-base theory can help. A Lewis Base is often the ligand of a coordination compound with the metal acting as the Lewis Acid (see Oxidation States of Transition   Metals ).

Al3++6H2O [⇌ Al(H2O)6]3+

The aluminum ion is the metal and is a cation with an unfilled valence shell, and it is a  Lewis Acid. Water has lone-pair electrons and is an anion, thus it is a Lewis Base.  

Figure 3 (Left): Aluminum ion acts as a lewis acid and accepts the electrons from water, which is acting as a lewis base. This helps explain the resulting hexaaquaaluminum(III) ion.

The Lewis Acid accepts the electrons from the Lewis Base which donates the electrons. Another case where Lewis acid-base theory can explain the resulting compound is the reaction of ammonia with Zn2+. 

Zn2++4NH3→[Zn(NH3)4]4+

Similarly, the Lewis Acid is the zinc Ion and the Lewis Base is NH3. Note how Brønsted Theory of Acids and Bases will not be able to explain how this reaction occurs because there are no H+ or OH− ions involved. Thus, Lewis Acid and Base Theory allows us to explain the formation of other species and complex ions which do not ordinarily contain hydronium or hydroxide ions. One is able to expand the definition of an acid and a base via the Lewis Acid and Base Theory. The lack of  H+ or OH− ions in many complex ions can make it harder to identify which species is an acid and which is a base. Therefore, by defining a species that donates an electron pair and a species that accepts an electron pair, the definition of a acid and base is expanded.

AmphoterismAs of now you should know that acids and bases are distinguished as two separate things however some substances can be both an acid and a base. You may have noticed this with water, which can act as both an acid or a base. This ability of water to do this makes it an amphoteric molecule. Water can act as an acid by donating its proton to the base and thus becoming its conjugate acid, OH-. However, water can also act as a base by accepting a proton from an acid to become its conjugate base, H3O+.

Water acting as an Acid:H2O+NH3→NH+4+OH−

Water acting as a Base:H2O+HCl→Cl−+H3O+

You may have noticed that the degree to which a molecule acts depends on the medium in which the molecule has been placed in. Water does not act as an acid in an acid medium and does not act as a base in a basic medium. Thus, the medium which a molecule is placed in has an effect on the properties of that molecule. Other molecules can also act as either an acid or a base. For example,

Al(OH)3+3H+→Al3++3H2O

where Al(OH)3 is acting as a Lewis Base. Al(OH)3+OH−→Al(OH)−4

where Al(OH)3 is acting as an Lewis Acid. Note how the amphoteric properties of the Al(OH)3 depends on what type of environment that molecule has been placed in

Arrhenius Concept of Acids and BasesThe Arrhenius definition can be summarized as "Arrhenius acids form hydrogen ions in aqueous solution with Arrhenius bases forming hydroxide ions."

IntroductionIn 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. As defined by Arrhenius, acid-base reactions are characterized by acids, which dissociate in aqueous solution to form hydrogen ions (H +) and bases, which form hydroxide (OH−) ions.Acids are defined as a compound or element that releases hydrogen (H+) ions into the solution.

In this reaction nitric acid (HNO3) disassociates into hydrogen (H+) and nitrate (NO3-) ions

when dissolved in water. Bases are defined as a compound or element that releases hydroxide (OH-) ions into the solution.

In this reaction lithium hydroxide (LiOH) dissociates into lithium (Li+) and hydroxide (OH-) ions when dissolved in water.

Definition of pH, pOH, "p", sample calculationsGeneral

The acid potential of aqueous solutions is measured in terms of the pH scale. The symbol "p" means take the negative logarithm of whatever follows in the formula. for pH, pOH, p[anything] . The pH scale is intended to be a convenience. Tremendous swings in hydrogen ion (hydronium ion) concentration occur in water when acids or bases are mixed with water. These changes can be as big as 1 x 1014, This means concentrations can change by multiples as big as one hundred trillion, 100,000,000,000,000.

The pH scale is a logarithmic scale. Every multiple of ten in H1+ concentration equals one unit on the logarithm scale. Physically the pH is intended to tell what the acid "potential" is for a solution.

In a sense the system is INVERTED so a low pH value indicates a great acid potential while a high pH indicates a low acid potential. ( Sad but true this is upside down and counter intuitive.) The pH values range from negative values to number above 14. Commonly the scale is often misrepresented as ranging from 0 to 14. We will see that negative values are possible.

 

Definition of pH, pOH, pKw, pKa, pKb

The p" factor" is defined as the log of the whatever quantity that follows the symbol. The "p" is an operator. It communicates the instruction to calculate the negative log of any quantity that follows the symbol. The definition of pH in equation form is

pH = -log[H1+] where [H1+] means the molar concentration of hydronium ions, M = moles / liter

This allows the definition of the following series of quantities.

pOH = -the negative log of the hydroxide ion molarity

log[OH-]pKw = -log Kw the negative log of the water ion product , KwpKa = -log Ka the negative log of the acid dissociation constant, KapKb = -log Kb the negative log of the base dissociation constant, Kb

The relationship pH + pOH = 14

In a water solution the ion product for water is:

[H+] [OH-] = Kw = 1 X 10-14

Take the -log of both sides of the equation

- log [H+] +(- log [OH- ]) = - log [1 X 10-14 ]

pH + pOH = 14

Calculations of pH

For strong acids like HCl the molar concentrations are essentially the hydronium ion concentration. These strong acids can produce solutions where the pH can be equal to or less than 1, the pH value would have a value from 0-14.

Example: Determination of pH from [H3O+]

What is the pH of a solution whose [H3O+] = 1 X 10-4 MpH = -log[H3O+]pH = - log[1 X 10-4]pH = - [ log 1 + log 10-4 ]Note: When you multiply numbers you always ADD their log forms

log 1 is always zero

log 10x = xso log 10-4 = -4

pH = - [ log 1 + log 10-4 ] = - [ 0 + (-4) ] = - [-4 ] = +4Example: Determination of pH from [H3O+] when coefficient is other than "1"

What is the pH of a solution whose [H3O+] = 2.5 X 10 -5 MpH = -log[H3O+]pH = - log[2.5 X 10 -5]pH = - [ log 2.5 + log 10-5 ]Note: When you multiply numbers you always ADD their log forms

log 10x = x so log 10-5 = -5

log 2.5 can be determined using a calculator having the log function key:

Enter the number in this case 2.5

depress the log key

Read the display which should be .3979 for this problempH = - [.3979 - 5] = 4.6021 or +4.602

Alternately if you can enter a number in scientific notation into your calculator key in 2.5 X 10 -5

depress the log keyRead the display which should be -4.602 for this problemMultiply by -1 to get + 4.602

Example: Determination of pH from [OH1- ] using defintion pOH and equation pH + pOH = 14

Calculate the pH of a solution that has a [OH1-] = 1 X 10-5 MDetermine pOH = -log[OH1- ] = -log [1 X 10-5 ] = 5Use the relationship pH + pOH = 14pH + 5 = 14pH = 14 -5 = 9

Hydrolysis of salts

Return to the Acid Base menu

A Brief Introduction to Hydrolysis Calculations

Hydrolysis happens when a substance chemically reacts with water. Hydrolysis should be distinguished from solvation, which is the process of water molecules associating themselves with individual solute molecules or ions.

I. Salts of Weak Acids

In general, all salts of weak acids behave the same, therefore we can use a generic salt to represent all salts of weak acids. Let NaA be a generic salt of a weak acid and A¯ its anion. Here are two specific examples of salts of weak acids:

Substance Formula The anion portion (A¯)sodium acetate NaC2H3O2 C2H3O2¯sodium benzoate C6H5COONa C6H5COO¯

By the way, the potassium ion, K+, (and several others) could also be used above without affecting any discussions of this topic. As a practical matter, only Na+ and K+ tend to get used in examples.

The generic chemical reaction (in net ionic form) for hydrolysis may be written thusly:

A¯ + H2O --> HA + OH¯

This reaction is of a salt of a weak acid (NOT the acid) undergoing hydrolysis, the name for a chemical reaction with water. The salt is NaAc and it is reacting with the water. Keep in mind that the acid (HAc) does not undergo hydrolysis, the salt does.

It is very important that you notice several things:

1) The Na+ (notice only OH¯ is written) IS NOT involved. Its source is the salt (NaA) that is dissolving in the water and it DOES NOT affect the pH. Its presence in both writing the chemical reactions and doing the calculations is deleted. However, keep in mind that Na+ is present in the solution. Some teacher might want to ask a "sneaky" question on a test.

2) HA is the UNDISSOCIATED acid. Keep in mind that it is not the acid that makes the acidic pH of a solution, it is the amount of hydrogen ion (or hydronium ion, H3O+, if you wish). In order to produce the hydrogen ion, the acid must dissociate.

3) There is free hydroxide ion (OH¯) in the solution!! This is the thing that makes the pH greater than 7.

Now, I can see a question forming in your mind. If there is acid (HA) and base (OH¯), why don't they just react and give back the reactants on the left side? Now, that really is a good question.

The answer? This reaction is an equilibrium. Now, if you are taking chemistry for the first time, you probably just got done with equilibrium a few weeks ago and it might have been hard to understand. That's understandable, but please realize that equilibrium is one of more important concepts in chemistry. Keep up the work!!

When a chemical reaction comes to equilibrium, there is a mixture of all involved substances in the reaction vessel. This mixture is characterized by a constant composition. (Keep in mind that constant composition DOES NOT imply equal composition.) The key point that makes a reaction come to equilibrium is that it is reversible. This means that both the forward reaction and the reverse reaction can happen, althought NOT initially with equal probability. The reaction comes to equilibrium when the rates of the two reactions (forward and reverse) become equal.

So, while it is true that the HA and OH¯ will react in the reverse direction, so can the A¯ and the H2O in the forward direction. The key point is that the reaction happens in such a way that a small amount (as opposed to zero) of HA and OH¯ are present at equilibrium.

When calculations are done, the important points will be (1) how much OH¯ is formed and (2) what is the pH of the solution?

Quick answers: (1) the amount of OH¯ formed will be greater than the 10¯7 M value present in pure water and (2) the pH will be greater than 7, so the solution of the salt of a weak acid will be basic.

II. Salts of Weak Bases

In general, all salts of weak bases behave the same, therefore we can use a generic salt to represent all salts of weak bases. Let B be a generic base and HB+ its salt. (Compare how this is worded compared to the "salt of weak acid" discussion.) HB+ is a cation, but that word is not used as much in discussions as is "anion" is above. Here are two specific examples of salts of weak bases:

Substance Formula The cation portion (HB+)ammonium chloride NH4Cl NH4

+

methyl ammonium nitrate CH3NH3NO3 CH3NH3+

The notation HB+ might be a bit confusing. Think of NH4+ this way:

HNH3+

NH3 is the base (symbolized by B) and an H+ has been attached to it in a chemical reaction. the NH3 has been protonated and the result (NH4

+) is now an acid. Why? Because it now has a proton to donate.

By the way, the chloride ion, Cl¯, and the nitrate ion, NO3¯ tend to be used in examples. Other anions of strong acids could be used, but their use is fairly uncommon.

The generic chemical reaction (in net ionic form) for hydrolysis reaction may be written thusly:

HB+ + H2O --> B + H3O+

This reaction is of a salt of a weak base (NOT the base) undergoing hydrolysis, the reaction with water. The salt in this case is HB+Cl¯ and it is reacting with the water. Remember, the most common specific example would be ammonium chloride (NH4

+Cl¯). Keep in mind that the base (generic example = B, specific example = ammonia or NH3) does not undergo hydrolysis, the salt does.

It is very important that you notice several things:

1) There is an anion involved, but it is usually not written. For example Cl¯ could be the anion, but it IS NOT involved. Its source is the salt (HB+Cl¯) that is dissolving in the water and it DOES NOT affect the pH. Its presence in writing the appropriate chemical reactions and doing the calculations is deleted. However, keep in mind that Cl¯ is present in the solution. Some teacher might want to ask a "sneaky" question on a test.

2) B is the UNPROTONATED base. Keep in mind that it is not the base that makes the basic pH of a solution, it is the amount of hydroxide ion (OH¯). In order to produce it, the base must protonated by the water.

3) There is free hydronium ion (H3O+) in the solution!! This is the thing that makes the pH less than 7.

Now, I can see a question forming in your mind. If there is base (B) and acid (H3O+), why don't they just react and give back the reactants on the left side? Now, that really is a good question.

The answer, of course, is given in above in the discussion of salts of weak acids. It would be the same explanation here, so I won't repeat it. What you might want to do, however, is look at the different phrasing in part I as compared to part II.

Of course, when calculations are done, the important points will be (1) how much H3O+ is formed and (2) what is the pH of the solution?

Quick answers: (1) the amount of H3O+ formed will be greater than the 10¯7 M value present in pure water and (2) the pH will be less than 7, so the solution of the salt of a weak base will be acidic.

saltCharacteristics and Classification of Salts

The most familiar salt is sodium chloride, the principal component of common table salt. Sodium chloride, NaCl, and water, H2O, are formed by neutralization of sodium hydroxide, NaOH, a base, with hydrogen chloride, HCl, an acid: HCl+NaOH→NaCl+H2O. Most salts are ionic compounds (see chemical bond); they are made up of ions rather than molecules. The chemical formula for an ionic salt is an empirical formula; it does not represent a molecule but shows the proportion of atoms of the elements that make up the salt. The formula for sodium chloride, NaCl, indicates that equal numbers of sodium and chlorine atoms combine to form the salt. In the reaction of sodium with chlorine, each sodium atom loses an electron, becoming positively charged, and each chlorine atom gains an electron, becoming negatively charged (see oxidation and reduction); there are equal numbers of positively charged sodium ions and negatively charged chloride ions in sodium chloride. The ions in a solid salt are usually arranged in a definite crystalline structure, each positive ion being associated with a fixed number of negative ions, and vice versa.A salt that has neither hydrogen (H) nor hydroxyl (OH) in its formula, e.g., sodium chloride (NaCl), is called a normal salt. A salt that has hydrogen in its formula, e.g., sodium bicarbonate (NaHCO3), is called an acid salt. A salt that has hydroxyl in its formula, e.g., basic lead nitrate (Pb[OH]NO3), is called a basic salt. Since a salt may react with a solvent to yield different ions than were present in the salt (see hydrolysis), a solution of a normal salt may be acidic or basic; e.g., trisodium phosphate, Na3PO4, dissolves in and reacts with water to form a basic solution.In addition to being classified as normal, acid, or basic, salts are categorized as simple salts, double salts, or complex salts. Simple salts, e.g., sodium chloride, contain only one kind of positive ion (other than the hydrogen ion in acid salts). Double salts contain two different positive ions, e.g., the mineral dolomite, or calcium magnesium carbonate, CaMg(CO3)2. Alums are a special kind of double salt. Complex salts, e.g., potassium ferricyanide, K3Fe(CN)6, contain a complex ion that does not dissociate in solution. A hydrate is a salt that includes water in its solid crystalline form; Glauber's salt and Epsom salts are hydrates.Salts are often grouped according to the negative ion they contain, e.g., bicarbonate or carbonate, chlorate, chloride, cyanide, fulminate, nitrate, phosphate, silicate, sulfate, or sulfide.