Acids, Bases and BuffersThe BrØnsted-Lowry definitions of an acid and a base are:

Acid: species that donates a protonBase: species that can accept a proton

When an acid is added to water, the acid dissociates releasing H+ ions into solution. Depending on the formula and bonding, different acids can release different numbers of protons: HCl H+ + Cl- - this is monoprotic as it releases one protonH2SO4 2H+ + SO4

2- - this is diprotic as it releases two protonsH3PO4 3H+ + PO4

3- - this is triprotic as is releases three protons

Writing ionic equations1. Write out the balanced chemical equation2. Split up each compound into its ions – be careful with solids – they don’t

dissociate, only aqueous solutions dissociate!3. Cancel species that appear on both sides of the equation Example: Aqueous sodium carbonate and hydrochloric acid

Na2CO3 + 2HCl 2NaCl + CO2 + H2O 2Na+ + CO3

2- + 2H+ + 2Cl- 2Na+ + 2Cl- + CO2 + H2O CO3

2- + 2H+ CO2 + H2O

Conjugate acid-base pairs transform into each other by the gain or loss of a proton.

Conjugate acid-base pairs are pairs of two species that transform into each other by gain or loss of a proton.

The ethanoic acid loses a proton to become an ethanoate ion. The ethanoate ion can accept a proton to become ethanoic acid. The ethanoic acid is the acid, the ethanoate ion is the base.

The water molecule can gain a proton to become H3O+. The H3O+ loses a proton to become water. The H3O+ is the acid, the H2O is the base.

Strong and Weak Acids

pH = - [H+] = [H+]eqm = [HA]initial Calculation 1: Strong Acid

A strong acid is a proton donor that completely dissociates in aqueous solution.

H2SO4 2H+ + SO42-

The value of Kc is extremely large, because equilibrium lies very far to the right.

CH3COOH CH3COO-

The value of Kc is small, because equilibrium lies to the left.

A weak acid is a proton donor that only partially dissociates in aqueous solution.

Calculating pH:

Strong acids are fully dissociated in aqueous solution, so we can say that the concentration of H+ is the same as the initial concentration of the acid itself.

A sample of HCl has a concentration of 1.22x10-3 moldm-3 .

What is it’s pH? HCl is a strong acid, therefore [H+] = [HCl][H+] = 1.22x10-3 moldm-3

pH = -log [H+] = -log (1.22x10-3 moldm-3 ) = 0.76

Calculation 2: Strong base using KwKw = [H+][OH-] [H+]=

pKa = -Ka=

Kw is known as the ionic product of water. Kw has a value of 1x10-14 at 25°. The definition of Kw is represented by the following equation: A strong base fully dissociates in aqueous solution. Therefore, we can say the concentration of OH- is the same as the initial concentration of the base.

We can use the Kw equation in rearranged form to find the [H+] and thereby find the pH.

A solution of KOH has a concentration of 0.050moldm-3. What is its pH? [OH-] = [KOH] = 0.050[H+] = = = 2x10-13

pH = -log [H+] = -log (2x10-3 ) = 12.70

The acid dissociation constant, Ka, shows the extent of the dissociation of an acid. It is given by the expression:

Ka : The Acid Dissociation Constant

𝐾 𝑎=¿¿ Ka values can be made more manageable if expressed in logarithmic form, called pKa :

Strong Acids have high Ka values and low pKa values – this indicates a large extent of dissociation.

Calculation 3: Weak Acid using KaKa= ¿The concentration of a sample of nitrous acid HNO2 is 0.055moldm-3. Ka = 4.7 x10-4. Calculate the pH. Ka= : 4.7x10-4 =

= 5.08x10-3pH = - = 2.29

Strong Acid, Strong Base Strong Acid, Weak Base

Weak Acid, Weak Base Weak Acid, Strong Base

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Acid

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The equivalence point of the titration is the point at which the volume of one solution has exactly reacted with the volume of the second solution. This matches the stoichiometry of the reaction taking place.

When the base is first added, there is only a very small increase in pH – the acid is in great excess.

Within 1-2cm3 of the equivalence point, the pH starts to increase more quickly – the acid is now only present in small excess.

The vertical section shows the very sharp increase in pH brought about by a very small further addition of base. The equivalence point is in the middle of this vertical section.

As further base is added, there is little additional change in pH, because the base is in great excess.

An indicator must be chosen for a titration so that the end point is as close as possible to the pH value of the titration’s equivalence point. A suitable indicator changes colour within the vertical section of the titration curve.

An indicator is a weak acid that is one colour in its acid form and a different colour in its conjugate base form. An indicator at its end point has equal amounts of acid and conjugate base present.

Buffer SolutionsA buffer solution is a system that minimises the pH change on addition of small amounts of acid or base. A buffer system is made from a weak acid and a salt of the weak acid – for example, methanoic acid and sodium methanoate. Alternatively, the weak acid could be partially neutralised by a an aqueous alkali such as NaOH, to give a solution containing a mixture of the salt and an excess of the weak acid.

How does a buffer system work?In the ethanoic acid/sodium ethanoate buffer system: The weak acid CH3COOH dissociates partially: CH3COOH H+ + CH3COO-

The salt (conjugate base), CH3COO-Na+ dissociates completely: CH3COO-Na+ CH3COO- + Na+

The equilibrium mixture formed contains a high concentration of the weak acid and its conjugate base. The resulting buffer solution contains large reservoirs of the weak acid and its conjugate base , both of which control pH. The overall principle of the buffer solution is that the weak acid removes added alkali and the conjugate base removes added acid.

On addition of acid:

[H+] increased

Conjugate base CH3COO- reacts with H+

Equilibrium shifts left removing most of added H+ ions.

On addition of alkali:

[OH-] increased

Added OH- reacts with H+ to form H2O

CH3COOH dissociates to form CH3COO- + H+. Equilibrium shifts right restoring most of H+ ions that have reacted.

Human Blood: Carbonic Acid-Hydrogencarbonate buffer system

Human blood plasma needs to have a pH of between 7.35 and 7.45. This is controlled by the Carbonic acid-Hydrogencarbonate buffer.

The weak acid H2CO3 partially dissociates.

Hydrogencarbonate, HCO3- acts a conjugate

base.

On addition of acid:

[H+] increased

Conjugate base HCO3- reacts with H+

forming H2CO3

Equilibrium shifts left removing most of added H+ ions.

On addition of alkali:

[OH-] increased

Added OH- reacts with H+ to form H2O

H2CO3 dissociates to form HCO3- + H+.

Equilibrium shifts right restoring most of H+ ions that have reacted.

Plenty of H2CO3 to make more H+ if [OH-] increases

Plenty of HCO3- to combine

with H+ if [H+] increases

H2CO3 HCO3- + H+

Calculation 4: pH of a buffer solution

Buffer Solutions

Calculate the pH of the following buffer solution: 50 cm3

of 1.0 moldm-3

methanoic acid, Ka = 1.78 x 10-4

moldm-3

; 20 cm3

of 1.0 mol dm-3

sodium methanoate.

Ka= Number moles acid = 1 x 50x10-3 = 0.05Number moles sodium methanoate = 1 x 20x10-3 = 0.02¿pH = - = -log (4.45x10-4) = 3.35

The pH of a buffer depends on the amount of dissociation, the temperature and the concentration ratio of the weak acid and its conjugate base. The expression used for pH of a buffer solution is:

Calculation 5: pH change of a buffer solutionHow would we calculate the pH change when 10cm3 of 0.1 moldm-3 NaOH is added to the buffer solution in the previous example?

What we already know about the buffer solution: Initial moles of acid = 0.05 molesInitial moles of NaCOOH = 0.02 molesInitial pH of buffer = 3.35Ka = 1.78 x 10-4 moldm-3

When NaOH is added, a reaction occurs between the acid and the NaOH base:

HCOOH + OH-

HCOO- + H2O

So, for every one mole of NaOH added, we get one mole more of the conjugate base, NaCOOH and one mole less of acid.

Initial Moles Change Final Moles

HCOOH 0.05 -0.001 0.049

NaCOOH 0.02 +0.001 0.021

Moles NaOH added: 0.1 x 10x10-3 = 0.001Therefore, the amount of acid present will go down by this amount and the amount of conjugate base will go up by this amount. We can now use a Ka

calculation with the new amounts present in the buffer solution to calculate the pH:

pH= -log(4.15x10-4

) = 3.38 therefore pH change = 3.38 – 3.35 = 0.03

Calculation 6: Neutralisation

1. Calculate moles of H+ before reaction.

2. Calculate moles of OH- before reaction

3. Calculate excess amount of H+ or OH- remaining after the reaction

4. Calculate excess concentration of OH- or H+

5. Find pH suing appropriate calculation.

Calculate the pH when 62cm3 of 1.05moldm-3 NaOH is added to 23cm3 of 0.91moldm-3 HCl.

1. Moles H+ = 0.91 x 23x10-3 = 0.02093 moles2. Moles OH- = 1.05 x 62x10-3 = 0.0651 moles3. OH- is in excess so moles of OH- remaining:

0.0651 – 0.02093= 0.04417 moles 4. [OH-] = = 0.5196moldm-3 (85cm3 is the total volume of

the two solutions mixed together)5. [H+] = = pH = - = -

= 13.72

Calculate the pH when 35cm3 of 1.25moldm-3 NaOH is added to 73cm3 of 0.92moldm-3 HCl.

1. Moles H+ = 0.92 x 73x10-3 = 0.06716 moles2. Moles OH- = 1.25 x 35x10-3 = 0.04375 moles3. H+ is in excess so moles of H+ remaining: 0.06716 – 0.04375= 0.02341 moles 4. [H+] = = 0.21676moldm-3

5. pH = -log10 (0.21676) = 0.66