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Introduction to Acids and BasesIntroduction to Acids and Bases

Pure or distilled water undergoes a veryslight ionization as shown below.

H2O(l) H+(aq) + OH-(aq)

The equilibrium constant for water is givenby:

Kw = [H+][OH-] = 1.00 x 10-14

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Note that the product of [H+] and [OH-] is aconstant at 25°C and that [H+] and [OH-]

areinversely proportional to each other.

The pH scale has been devised to determine

the molarity of the hydrogen ion, [H+], in anaqueous solution and is given by:

pH = -log[H+]

Similarly, pOH is defined as:

pOH = -log[OH-]

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Another useful equation is:

pH + pOH = 14.00

It follows that:

[H+] = 10-pH and [OH-] = 10-pOH

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Acidic, basic, or neutral solutions can bedistinguished as shown below:

Solution pH [H+] pOH [OH-]

Acidic < 7.00 > 10-7 > 7.00 < 10-7

Neutral = 7.00 = 10-7 = 7.00 = 10-7

Basic > 7.00 < 10-7 < 7.00 > 10-7

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An alternative approach to the relationshipbetween pH and pOH is shown below.

pOH

14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

pH

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Increasing Acidity Decreasing Acidity

Decreasing AcidityIncreasing AcidityN

eu

tral

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Strong AcidsStrong Acids

There are six strong acids.

HCl, HNO3, HClO4, HI, HBr, and H2SO4

When given any of the above acids

always assume 100% ionization.

100% ionization is shown by using a single arrow in the ionization equation.

The ionization equation for H2SO4 is not

intuitively obvious to the most casual observer.

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All the strong acids are assumed to have

one ionizable hydrogen including H2SO4.

The ionization of H2SO4 must be shown as:

H2SO4(aq) → H+(aq) + HSO4-(aq)

The single arrow as always indicates that the ionization is 100% and there

are no H2SO4 molecules remaining in solution.

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Strong Acid ProblemStrong Acid Problem

Determine the pH and pOH of a 0.20 M HClsolution.

[HCl] = 0.20 M

HCl(aq) → H+(aq) + Cl-(aq)

pH = -log[H+] = -log(0.20) = 0.70

pH + pOH = 14.00

pOH = 14.00 – 0.70 = 13.30

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Another Strong Acid ProblemAnother Strong Acid Problem

Determine the pH and pOH of a 10-10 M HNO3

solution.

[HNO3] = 10-10 M

HNO3(aq) → H+(aq) + NO3-(aq)

pH = -log[H+] = -log(10-10) = 10

What??

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How can a strong acid have a pHcorresponding to a base?

The answer lies with the autoionization ofwater.

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

[ ]I 0 0[ ]c +1.0 x 10-7 +1.0 x

10-7

[ ]e 1.0 x 10-7 1.0 x 10-

7

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Although water ionizes only to a slight extent,there is a dynamic equilibrium which remainsintact.

Distilled or pure water has a pH = 7 becausethe [H+] = [OH-] = 1.0 x 10-7 M.

[H+]T = [H+]water + [H+]nitric acid

[H+]T = 1.0 x 10-7 M + 10-10 M ≈ 1.0 x 10-7 M

pH = -log[H+] = -log(1.0 x 10-7) = 7.00

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Weak Acids Equilibrium ConstantWeak Acids Equilibrium Constant

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

[ ]I 0 0[ ]c +1.0 x 10-7 +1.0 x 10-7

[ ]e 1.0 x 10-7 1.0 x 10-7

Keq = [H+] [OH-]

Kw = Keq = [H+] [OH-] = 1.00 × 10-14

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For the reaction

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

remember the following:

This is an example of a heterogeneous equilibria because more than one

phase (state) is present.

When the reactants/products are solids or liquids, their concentration are

nearly constant and incorporated into the Keq.

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Pure solids and liquids are not incorporated into the equilibrium expression.

The position of the equilibrium is independent of the amount of solid or liquid present.

The solvent for the reaction does not appear in the equilibrium expression.

Ions or molecules appear as their molar

concentrations.

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The effect of temperature on the Keq

depends on which reaction is endothermic.

The inverse relationship between H+ and

OH- in any solution is given by Kw.

Remember the numerical value for Kw is

both reaction and temperature dependent.

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A Weak Acid ProblemA Weak Acid Problem

Acetic acid, HC2H3O2, has a Ka = 1.8 x 10-5.Calculate the hydrogen ion concentration in asolution prepared by adding 2.0 moles ofacetic acid to form one liter of solution.

Ka = 1.8 x 10-5 n = 2.0 mol HC2H3O2 V = 1.0 L

[HC2H3O2] =

[HC2H3O2] =

nV

2.0 mol HC2H3O2

1.0 L= 2.0 M

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HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

[ ]I 2.0 0 0[ ]c -x +x +x[ ]e 2.0 - x x x

Ka =[H+] [C2H3O2

-][HC2H3O2]

1.8 x 10-5 =2.0 - xx x×

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At this point in the problem an approximation

will be made called the 5% rule.

You will note that Ka = 1.8 x 10-5 is a very

small number and that x may be very small compared to the initial concentration of the acid.

This assumption will not always be true and should always be tested after solving for x which is the [H+].

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The approximation made will be 2.0 – x ≈ 2.0

which avoids using the quadratic formula.

x = [H+] = 6.0 x 10-3 M

The 5% rule is given by:

1.8 x 10-5 =2.0 - xx x× ≈ x2

2.0

%ion =[H+]

[HC2H3O2]× 100%

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%ion =

The 0.30% is well within the tolerance of 5%so the assumption is valid.

If the % ionization was more than 5%, then the

assumption would not be valid and you wouldhave to use the quadratic formula.

There are quadratic formula programs for theTI calculator and also the Solver function onthe TI.

6.0 x 10-3 M

2.0 M× 100% = 0.30%

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To learn more about the Solver, visit

http://www2.ohlone.edu/people2/joconnell/ti/solver8384.pdf

Another useful quantity that is used is the pKa

of an acid and is defined as pKa = -log Ka.

So for the current problem, acetic acid,

Ka = 1.8 x 10-5 and pKa = 4.74

pKa’s are more convenient to work with than

Ka’s.

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Combining Acidic SolutionsCombining Acidic Solutions

What is the pH of a solution obtained bymixing 235 mL of 0.0245 M HNO3 with 554 mL of 0.438 M HClO4?

[HNO3] = 0.0245 M [HClO4] = 0.438 MV1 = 235 mL V2 = 554 mL

HNO3(aq) → H+(aq) + NO3-(aq)

HClO4(aq) → H+(aq) + ClO4-(aq)

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[HNO3] = n/V

1 mol HNO3

n1=0.0245 mol HNO3

1.00 L× 235 mL ×

1 L

103 mL×

×1 mol H+

=5.76 × 10-3 mol H+

1 mol HNO3

n2=0.438 mol HClO4

1.00 L× 554 mL ×

1 L

103 mL×

×1 mol H+

=2.42 × 10-1 mol H+

1 mol HClO4

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.

[H+] =nT

VT

[H+] =2.48 × 10-1 mol H+

789 mL × 1 L103 mL

= 0.314 M

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Polyprotic AcidsPolyprotic Acids

Polyprotic acids are those acids, H2CO3 andH2SO4, which have more than one ionizablehydrogen.

The Ka value for each hydrogen is differentand Ka vastly decreases with each

hydrogen.

The first hydrogen is the easiest to removebecause the original molecule or ion is thestrongest acid.

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The reason for the increased difficultly isbecause each successive hydrogen is tryingto be removed from a more negative

species.

Compare the two ionization equations forcarbonic acid.

H2CO3(aq) H+(aq) + HCO3-(aq)

HCO3-(aq) H+(aq) + CO3

2-(aq)

The greater attraction between the HCO3

- and the H+ will make it more difficult to ionize.

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The pH of a diprotic or a triprotic acid isdetermined almost completely by the firstionization.

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Diprotic Acid ProblemDiprotic Acid Problem

Calculate the pH of a 0.0020 M solution ofcarbonic acid.

[H2CO3] = 0.0020 M

Ka1 = 4.4 x 10-7 Ka2 = 4.7 x 10-11

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H2CO3(aq) H+(aq) + HCO3-(aq)

[ ]I 0.0020 0 0[ ]c -x +x +x[ ]e 0.0020 - x x x

4.4 x 10-7 =0.0020 - x

x x×

Ka1 =[H+] [HCO3

-][H2CO3]

4.4 x 10-7 =0.0020 - x

x x×

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We will make the assumption that 0.0020 – x ≈

0.0020 and verify the assumption after thecalculation is made.

x = [H+] = [HCO3-] = 3.0 x 10-5 M

4.4 x 10-7 =0.0020

x2

%ion =[H+]

[HCO3-]

× 100%

%ion =3.0 x 10-5 M

2.0 x 10-3 M×100% =1.5%

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The %ion = 1.5% is well within the 5%,therefore the assumption is valid.

HCO3-(aq) H+(aq) + CO3

2-(aq)

[ ]I 3.0 x 10-5 3.0 x 10-5 0

[ ]c -x +x +x[ ]e 3.0 x 10-5-x 3.0 x 10-5+x x

•

You must use the concentrations of H+ andHCO3

- resulting from the first hydrogen ionizing.

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B

Ka2 =[H+] [CO3

2-][HCO3

-]

4.7 x 10-11=(3.0 x 10-5 + x) x×

(3.0 x 10-5 - x)

The extremely small value of the exponent,10-11, allows for the assumption3.0 x 10-5 ± x ≈ 3.0 x 10-5 which simplifies theabove expression to

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x = [H+] = [CO32-] = 4.7 x 10-11 M

%ion =[H+]

[HCO3-]

× 100%

%ion =4.7 x 10-11 M

3.0 x 10-5 M×100% =1.6 x 10-4%

The %ion = 1.6 x 10-4 % is well within the 5%,therefore the assumption is valid.

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What was the point of doing this rather longand tedious problem?

These steps can be applied to any diprotic or

triprotic acid and what you verify is that the[H+] contribution from the second or thirdhydrogen can be ignored.

So, what is the pH of a 0.0020 M solution ofcarbonic acid?

pH = -log[H+] = -log(3.0 x 10-5) = 4.52

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Common Ion EquilibriaCommon Ion Equilibria

Calculate the pH of a solution containing0.075 M nitrous acid and 0.20 M sodiumnitrite.

Ka = 4.5 x 10-4

[NaNO2] = 0.20 M [HNO2] = 0.075 M

NaNO2(aq) → Na+(aq) + NO2-(aq)

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HNO2(aq) H+(aq) + NO2-(aq)

[ ]I 0.075 0 0.20[ ]c -x +x +x[ ]e 0.075 - x x x

Ka =[H+] [NO2

-][HNO2]

4.5 x 10-4 =0.075 - x

x (0.20 + x)× ≈0.075 0.20x

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[H+] = 1.7 x 10-4 M

%ion =[H+]

[HNO2]× 100%

%ion =1.7 x 10-4 M

× 100%0.075 M

= 0.23%

The 5% rule applies so the previous twoassumptions are valid.

pH = -log[H+] = -log(1.7 x 10-4) = 3.77