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Additional Aqueous Additional Aqueous EquilibriaEquilibria

Chapter 17Chapter 17

Lawrence J. Henderson1878-1942.Discovered how acid-baseequilibria are maintainedin nature by carbonic acid/bicarbonate system in theblood. Developed bufferequation.

Karl A. Hasselbalch 1874-1962Developed logarithmic form of buffer equation.

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Use this “decision tree” to calculate pHUse this “decision tree” to calculate pHvalues of solutions of specific solutions.values of solutions of specific solutions.

1. Is it pure water? If yes, pH = 7.00.2. Is it a strong acid? If yes, pH = -log[HZ]3. Is it a strong base? If yes, pOH = -log[MOH] or pOH = -log (2 x [M(OH)2]) 4. Is it a weak acid? If yes, use the relationship Ka = x2/(HZ – x), where x = [H+]

5. Is it a weak base? If yes, use the relationship Kb = x2/(base – x), where x = [OH-]6. Is it a salt (MZ)? If yes, then decide if it is neutral, acid, or base; calculate its K value by the relationship KaKb = Kw, where Ka and Kb are for a conjugate system; then treat it as a weak acid or base.7. Is it a mixture of a weak acid and its weak conjugate base? It is a buffer; use the Buffer Equation.

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Common Ion EffectSo far, we’ve looked at solutions of weak acidsand solutions of weak bases.

Weak acid equilibrium: HX H+ + X- (described by Ka)

Weak base equilibrium: X- + H2O HX+ OH- (described by Kb)

What if you had both HX and X- in the same solution?

This could be obtained by adding some Na+X- salt to a solution of HX. The result is called a buffered solution.A buffered solution resists changes to pH when an acid or base is added.

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Buffered SolutionsBuffered SolutionsA buffer consists of a mixture of a weak acid (HX) and its conjugate base (X-).

An example of preparing a buffered solution: Add 0.10 mole of lactic acid (H-Z) and 0.14 mole of sodium lactate (Na-Z) to a liter of solution.A buffer resists a change in pH when a small amount of OH- or H+ is added.The reason the buffered solution resists pH change becomes clear when we remember LeChâtlier’s principle.

HX(aq) H+(aq) + X-(aq)

LeChâtlier says the above equation will (1) shift to the left if we add HCl or (2) shift to the right if we add NaOH -- thus resisting a change in [H+]

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Buffered SolutionsBuffered Solutions

][[X-]

[HX]KH a

][X

[HX]loglogK]log[H a

We can predict the pH of a buffered solution:

Taking –logs of both sides and rearranging:

][X

[HX]logpKpH a or

[HX]

][XlogpKpH a

Henderson-Hasselbalch equation or Buffer equation[acid]

[base]logpKpH a

[HX]

[H+][X-]Ka

Rearranging,

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Buffered SolutionsBuffered SolutionsWhat is the pH of a buffered solution of 0.2 M HF and 0.1 M NaF?(buffer is weak acid HF and its conj. base (salt) F-)

Use:[HF]

][FlogpKpH a

For HF, pKa = -log Ka = -log(6.8 x 10-4) = 3.16

So, 2.860.303.160.2

0.1log3.16pH

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One often wants a buffer to have a specific [base]/[acid] ratio.The H2CO3/HCO3

- buffer in blood should be at a pH=7.40.What must the [base]/[acid] ratio be to obtain this?

For H2CO3 H+ + HCO3- Ka = 4.3x10-7 pKa = 6.37

Henderson-Hasselbalch:

[acid]

[base]logpKpH a

Rearrange and solve for [base]/[acid]:

03.137.640.7pKpH[acid]

[base]log a

1

11

]CO[H

][HCO

[acid]

[base]

32

3

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Adding strong acids and strong bases to buffersAssume 1 L of a lactic acid/lactate buffer, [HLac] = 0.061 M; and [Lac-1] = 0.079 M.The pKa for lactic acid is 3.85.

The pH is pKa + log[(0.079)/(0.061)] = 3.96

Suppose you add 0.020 mol H+ (strong acid) to this buffer with no change in volume. Then, there will be a neutralization reaction between H+ and Lac-

H+ + Lac- HLac

init.

change -0.020M -0.020M +0.020 M

after rxn 0 0.059 M 0.081 MHenderson-Hasselbalch:

[HLac]

][LaclogpKpH

-

a 71.3)14.0(85.3081.0

059.0log85.3

0.020M 0.079M 0.061M

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Adding strong acids and strong bases to buffers

Assume 1 L of the lactic acid/lactate buffer from earlier slide. [HLac] = 0.061 M; and [Lac-1] = 0.079 MThe pKa for lactic acid is 3.85.

Suppose you add 0.020 mol OH- (strong base) to this buffer with no change in volume. Then, there will be a neutralization reaction between OH- and HLac

OH- + HLac Lac-

init.

change

after rxn

0.020M

-0.020M -0.020M +0.020 M0.061M 0.079M

0 0.041 M 0.099 MHenderson-Hasselbalch:

[HLac]

][LaclogpKpH

-

a 23.4)+0.38(85.3041.0099.0

log85.3

The pH is pKa + log[(0.079)/(0.061)] = 3.96

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Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations

buret(with 0.10 M OH- (aq)

beaker(with 0.10 M H+ (aq)& indicator)

V=50.0 mL

Titration curve

HCl + NaOH NaCl

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Acid-Base TitrationsAcid-Base Titrations

Strong Base – Weak Acid Titration

Half-way through the titration,the pH = pKa = 4.74

At the end of the titration,the pH is determined by the concentration of [NaAc] = 0.050 M.The pH = 8.72

HAc + NaOH NaAc

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Acid-Base TitrationsAcid-Base Titrations Titrations of Polyprotic AcidsTitrations of Polyprotic Acids• In polyprotic acids, each ionizable proton dissociates

in steps.

• In the titration of H2CO3 with NaOH there are two equivalence points:

–one for the formation of HCO3-

H2CO3 + OH-→ HCO3- + H2O

–one for the formation of CO32-

HCO3- + OH- → CO3

2- + H2O

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Solubility EquilibriaSolubility Equilibria Solubility-Product Constant, KSolubility-Product Constant, Kspsp

• Consider

BaSO4(s) Ba2+(aq) + SO42-(aq)

]][SO[BaK -24

2 = Ksp

•Ksp is the solubility product. (Remember, BaSO4 is

ignored because it is a pure solid so its concentration is constant.) The larger the solubility product, the more soluble the salt.

•for which

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Solubility EquilibriaSolubility Equilibria Solubility-Product Constant, KSolubility-Product Constant, Kspsp

• In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers.

Examples: for Ag2CO3 Ksp = [Ag+]2[[CO3-2]

for Al2(SO4)3 is Ksp = [Al+3]2[SO4-2]3

• Solubility is the amount (grams) of substance that dissolves to form a saturated solution.

• Molar solubility (s) is the number of moles of solute dissolving to form a liter of saturated solution.

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Solubility and KSolubility and Kspsp

Problem solving – always “let s = molar solubility” Two types of problems: (1)Calculate Ksp from solubility(2)Calculate solubility from Ksp

Problem examples:(a) Calculate Ksp for PbS, if the solubility = 1.73 x 10-14 M

Let s = molar solubility = [PbS(aq)] = 1.73 x 10-14

[Pb+2] = s = 1.73 x 10-14

[S-2] = s = 1.73 x 10-14

Ksp = [Pb+2][S-2] = s2 = (1.73 x 10-14)2 = 2.95 x 10-28

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(b) Calculate Ksp for Co(OH)2 ; solubility = 6.88 x 10-6 M

Let s = molar solubility = [Co(OH)2(aq)] = 6.88 x 10-6

[Co+2] = s = 6.88 x 10-6

[OH-] = 2s = 2(6.88 x 10-6) = 1.38 x 10-5

Ksp = [Co+2][OH-]2 = (6.88 x 10-6)(1.38 x 10-5)2

= 1.31 x 10-15

[Alternatively, Ksp = 4s3 = 4(6.88 x 10-6)3 = 1.30 x 10-15]

(c) Calculate solubility of CdS; Ksp = 8 x 10-28

Let s = molar solubility (unknown) = [CdS(aq)]Ksp = [Cd+2][S-2] = (s)(s) = s2 = 8 x 10-28

s = 3 x 10-14

(d) Calculate solubility of CaF2; Ksp = 3.9 x 10-11

Let s = molar solubility (unknown) = [CaF2(aq)]

Ksp = [Ca+2][F-]2 = (s)(2s)2 = 4s3 = 3.9 x 10-11

s = 2.1 x 10-4

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Factors That Affect SolubilityFactors That Affect Solubility Common-Ion EffectCommon-Ion Effect• Solubility is decreased when a common ion is added.• This is an application of Le Châtelier’s principle:

CaF2(s) Ca2+(aq) + 2F-(aq)

• As F- (from NaF, say) is added, the equilibrium shifts to the left.

•Therefore, CaF2(s) is formed and precipitation occurs.

•As NaF is added to the system, the solubility of CaF2 decreases.

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Common-Ion EffectCommon-Ion Effect

What is solubility of AgCl in a 0.1 M solution of NaCl (which contains Cl-, a common ion)?

Example problem. What is solubility of AgCl with and without NaCl?Given: Ksp = 1.8 x 10-10 for AgClLet s = solubility of AgCl (unknown) = [AgCl(aq)] Ksp = 1.8 x 10-10 = s2 Solubility = s = (1.8 x 10-10 )1/2 = 1.35 x 10-5 M (without NaCl)

AgCl (s) Ag+(aq) + Cl- (aq) x x+0.1

Ksp = [Ag+][Cl-] = (s)(s+0.1) (s)(0.1) = 1.8 x 10-10

s = 1.8 x 10-9 M (with NaCl)

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Solubility and pHSolubility and pHHow would we increase the solubility of a sparingly soluble salt such as CaF2?

In water, the equilibrium is: CaF2(s) Ca2+(aq) + 2 F- (aq)LeChatelier says that to increase solubility, we must remove F- ions.If we add a strong acid, H+, this will cause the reaction: H+(aq) + F- (aq) HF (aq)

(Remember, HF is a weak acid)

This reaction removes F- ions thus driving equilibrium in the top equation to the right which means more CaF2 dissolves.

As acid is added, pH decreases and solubility of CaF2 increases.