Buffers and the Common-Ion Effect (Ch 16) a Buffer Works Through the Common-Ion Effect

Buffers and the Common-ion Effect (Ch 16) A buffer works through the common-ion effect.

Acetic acid in water dissociates slightly to produce some acetate ion:

- + CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) acetic acid acetate ion

If NaCH3COO is added, it provides a source of - CH3COO ion, and the equilibrium shifts to the left. - CH3COO is common to both solutions.

- * + [CH3COOH]init [CH3COO ]added % Dissociation [H3O ] pH

0.10 0.00 1.3 1.3x10-3 2.89

0.10 0.050 0.036 3.6x10-5 4.44

0.10 0.10 0.018 1.8x10-5 4.74

0.10 0.15 0.012 1.2x10-5 4.92

[CH COOH] * % Dissociation = 3 dissoc x 100 [CH3COOH]init

The buffer components (HA and A-) are able to consume - + small amounts of added OH or H3O by a shift in equilibrium position.

- + CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq)

- + Added OH reacts with Added H3O reacts with - CH3COOH, causing a shift to CH3COO , causing a the right. shift to the left.

The shift in equilibrium position absorbs the change in + - [H3O ] or [OH ], and the pH changes only slightly.

Buffer has more HA after Buffer has equal Buffer has more A- after + - - addition of H3O . concentrations of A and HA. addition of OH .

+ - H3O OH

+ - - - H2O + CH3COOH ← H3O + CH3COO CH3COOH + OH → CH3COO + H2O

A 100-mL sample of The addition of 1 mL of strong acid (left) dilute HCl is adjusted or strong base (right) changes the pH by to pH 5.00. several units.

A 100-mL sample of The addition of 1 mL of strong acid (left) an acetate buffer is or strong base (right) changes the pH very adjusted to pH 5.00. little.

The acetate buffer is made by mixing 1 M CH3COOH ( a weak acid) with - 1 M CH3COONa (which provides the conjugate base, CH3COO ).

- + CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) [CH COO-][H O+] [CH COOH] K = 3 3 [H O+] = K x 3 a 3 a - [CH3COOH] [CH3COO ]

+ Since Ka is constant, the [H3O ] of the solution depends on the ratio of buffer component concentrations. [HA] If the ratio increases, [H O+] increases. [A-] 3 [HA] If the ratio decreases, [H O+] decreases. [A-] 3

©2013 McGraw-Hill Ryerson Limited 17-7 + Sample Problem 17.1 Calculating the Effect of Added H3O or OH- on Buffer pH PROBLEM: Calculate the following pH values(Ka of CH3COOH = 1.8 x 10-5 , assume the additions cause a negligible change in volume.)

(a) pH of a buffer solution consisting of 0.50 mol/L CH3COOH and 0.50 mol/L CH3COONa (b) pH after adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution (c) pH after adding 0.020 mol of HCl to 1.0 L of the buffer solution in (a).

- PLAN: We can calculate [CH3COOH]init and [CH3COO ]init from the given information. From this we can find the starting pH. For (b) and (c) - + we assume that the added OH or H3O reacts completely with the buffer components. We write a balanced equation in each case, set up + a reaction table, and calculate the new [H3O ].

©2013 McGraw-Hill Ryerson Limited 17-8 Sample Problem 17.1 SOLUTION: (a) - + Concentration(mol/L) CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) Initial 0.50 - 0.50 0 Change −x - +x +x Equilibrium 0.50 - x - 0.50 + x x

Since Ka is small, x is small, so we assume - [CH3COOH] = 0.50 – x ≈ 0.50 mol/L and [CH3COO ] = 0.50 + x ≈ 0.50 mol/L

+ [CH3COOH] -5 0.50 -5 x = [H3O ] = Ka x ≈ 1.8x10 x = 1.8x10 mol/L - [CH3COO ] 0.50 -5 Checking the assumption: pH = -log(1.8x10 ) = 4.74 -5 1.8x10 mol/L x 100 = 3.6x10-3% (< 5%; assumption is justified.) 0.50 mol/L

− 0.020 mol − (b) [OH ]added = = 0.020 M OH 1.0 L soln Setting up a reaction table for the stoichiometry: - - Concentration (mol/L) CH3COOH(aq) + OH (aq) → CH3COO (aq) + H2O(l) Initial 0.50 0.020 0.50 - Change -0.020 -0.020 +0.020 - Equilibrium 0.48 0 0.52 -

Setting up a reaction table for the acid dissociation, using new initial [ ]:

- + Concentration (mol/L) CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) Initial 0.48 - 0.52 0 Change −x - +x +x Equilibrium 0.48 - x - 0.52 + x x

Since Ka is small, x is small, so we assume - [CH3COOH] = 0.48 – x ≈ 0.48 mol/L and [CH3COO ] = 0.52 + x ≈ 0.52 mol/L

+ [CH3COOH] -5 0.48 -5 x = [H3O ] = Ka x ≈ 1.8x10 x = 1.7x10 mol/L - [CH3COO ] 0.52 pH = -log(1.7x10-5) = 4.77

+ 0.020 mol + (c) [H3O ]added = = 0.020 M H3O 1.0 L soln Setting up a reaction table for the stoichiometry: - + Concentration (mol/L) CH3COO (aq) + H3O (aq) → CH3COOH(aq) + H2O(l) Initial 0.50 0.020 0.50 - Change -0.020 -0.020 +0.020 - Equilibrium 0.48 0 0.52 -

Setting up a reaction table for the acid dissociation, using new initial [ ]:

- + Concentration(mol/L) CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) Initial 0.52 - 0.48 0 Change −x - +x +x Equilibrium 0.52 - x - 0.48 + x x

Since Ka is small, x is small, so we assume - [CH3COOH] = 0.52 – x ≈ 0.52 mol/L and [CH3COO ] = 0.48 + x ≈ 0.48 mol/L

+ [CH3COOH] -5 0.52 -5 x = [H3O ] = Ka x ≈ 1.8x10 x = 2.0x10 mol/L - [CH3COO ] 0.48 pH = -log(2.0x10-5) = 4.70

- + HA(aq) + H2O(l) A (aq) + H3O (aq)

+ - [H3O ][A ] + [HA] K = [H3O ] = Ka x a [HA] [A-]

[HA] -log[H O+] = -logK – log 3 a [A-]

[base] pH = pK + log a [acid]

The buffer capacity is a measure of the “strength” of the buffer, its ability to maintain the pH following addition of strong acid or base.

The greater the concentrations of the buffer components, the greater its capacity to resist pH changes. The closer the component concentrations are to each other, the greater the buffer capacity.

When strong base is added, the pH increases least for the most concentrated buffer.

The buffer range is the pH range over which the buffer is effective. Buffer range is related to the ratio of buffer component concentrations. [HA] The closer is to 1, the more effective the buffer. [A-]

If one component is more than 10 times the other, buffering action is poor. Since log10 = 1, buffers have a usable

range within ± 1 pH unit of the pKa of the acid component.

PROBLEM: The molecular scenes below represent samples of four HA/A- buffers. (HA is blue and green, A- is green, and other ions and water are not shown.)

(a) Which buffer has the highest pH? (b) Which buffer has the greatest capacity? (c) Should we add a small amount of concentrated strong acid or strong base to convert sample 1 to sample 2 (assuming no volume changes)?

PLAN: Since the volumes of the solutions are equal, the scenes represent molarities as well as numbers. We count the particles of each species present in each scene and calculate the ratio of the buffer components.

SOLUTION: [A-]/[HA] ratios: sample 1, 3/3 = 1; sample 2, 2/4 = 0.5; sample 3, 4/4 = 1; and sample 4, 4/2 = 2.

(a) As the pH rises, more HA will be converted to A-. The scene with the highest [A-]/[HA] ratio is at the highest pH. Sample 4 has the highest pH because it has the highest ratio.

(b) The buffer with the greatest capacity is the one with the [A-]/[HA] closest to 1. Sample 3 has the greatest buffer capacity.

(c) Sample 2 has a lower [A-]/[HA] ratio than sample 1, so we need to increase the [A-] and decrease the [HA]. This is achieved by adding strong acid to sample 1.

• Choose the conjugate acid-base pair.

– The pKa of the weak acid component should be close to the desired pH. • Calculate the ratio of buffer component concentrations.

[base] pH = pK + log a [acid] • Determine the buffer concentration, and calculate the required volume of stock solutions and/or masses of components. • Mix the solution and correct the pH.

PROBLEM: An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of the acid rain on

limsetone-rich soils. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 mol/L NaHCO3 - -11 to make the buffer? Ka of HCO3 is 4.7x10 .

- 2- PLAN: The conjugate pair is HCO3 (acid) and CO3 (base), and we - know both the buffer volume and the concentration of HCO3 . We can calculate the ratio of components that gives a pH of

10.00, and hence the mass of Na2CO3 that must be added to make 1.5 L of solution.

SOLUTION: + -pH -10.00 -10 [H3O ] = 10 = 10 = 1.0x10 mol/L [CO 2-][H O+] HCO -(aq) + H O(l) H O+(aq) + CO 2-(aq) 3 3 3 2 3 3 Ka = - [HCO3 ] ©2013 McGraw-Hill Ryerson Limited 17-22 Sample Problem 17.3 Preparing a Buffer

K [HCO -] (4.7x10-11)(0.20) [CO 2-] = a 3 = = 0.094 mol/L 3 + -10 [H3O ] 1.0x10 2- 0.094 mol CO 2- Amount (mol) of CO3 needed = 1.5 L soln x 3 1 L soln

2- = 0.14 mol CO3

0.14 mol Na CO x 105.99 g Na2CO3 2 3 = 15 g Na2CO3 1 mol Na2CO3

The chemist should dissolve 15 g Na2CO3 in about 1.3 L of 0.20 mol/

L NaHCO3 and add more 0.20 mol/L NaHCO3 to make 1.5 L. Using a pH meter, she can then adjust the pH to 10.00 by dropwise addition of concentrated strong acid or base.

PROBLEM: A first year university student working in a lab is asked to prepare a buffer by adding 5.00mL of 5.0 mol/L NaOH To 100.0 mL of 0.50 mol/L HCOOH. What will be the pH of -4 the resulting buffer solution?(Ka HCOOH= 1.8 ×10 )

PLAN: Knowing the amount of strong base(NaOH) and the weak acid (HCOOH), we can calculate the remaining amount of weak acid - and the formed conjugate base(HCOO ) in solution. Knowing Ka Of the acid and using Henderson Hasselbalch equation, we can find the pH of the buffer in solution

SOLUTION: + - HCOOH(aq) + NaOH(aq) → Na HCOO (aq) + H2O(l) After dilution,

5.0 mol/L x 5.00 mL [NaOH]initial = = 0.238 mol/L 105.0 mL

0.50 mol/L x 100.0 mL [HCOOH]initial = = 0. 476 mol/L 105.0 mL

+ - HCOOH(aq) + NaOH(aq) → Na HCOO (aq) + H2O(l)

Initial 0.476 0.238 0 -

Change − 0.238 − 0.238 +0.238 -

Equilibrium +0.238 0 +0.238 -

- mol/L [HCOO ] -4 0.238 pH = pKa + log = -log 1.8 ×10 + log [HCOOH] 0.238 mol/L

= 3.74

An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In-).

- + The ratio [HIn]/[In ] is governed by the [H3O ] of the solution. Indicators can therefore be used to monitor the pH change during an acid-base reaction.

The color of an indicator changes over a specific, narrow pH range, a range of about 2 pH units.

pH < 6.0 pH = 6.0-7.5 pH > 7.5

In an acid-base titration, the concentration of an acid (or a base) is determined by neutralizing the acid (or base) with a solution of base (or acid) of known concentration.

The equivalence point of the reaction occurs when the number of moles of OH- added equals the number of + moles of H3O originally present, or vice versa.

The end point occurs when the indicator changes color. - The indicator should be selected so that its color change occurs at a pH close to that of the equivalence point.

The pH increases gradually when excess base has been added.

The pH rises very rapidly at the equivalence point, which occurs at pH = 7.00.

The initial pH is low.

strong acid–strong base titration

Initial pH

+ [H3O ] = [HA]init + pH = -log[H3O ]

pH before equivalence point

+ initial mol H3O = Vacid x Macid - mol OH added = Vbase x Mbase + + - mol H3O remaining = (mol H3O init) – (mol OH added) + + mol H3O remaining + [H3O ] = pH = -log[H3O ] Vacid + Vbase

strong acid–strong base titration

pH at the equivalence point pH = 7.00 for a strong acid-strong base titration.

pH beyond the equivalence point

+ initial mol H3O = Vacid x Macid - mol OH added = Vbase x Mbase - - + mol OH excess = (mol OH added) – (mol H3O init) mol OH- [OH-] = excess

Vacid + Vbase pOH = -log[OH-] and pH = 14.00 - pOH

40.00 mL of 0.1000 mol/L HCl is titrated with 0.1000 mol/L NaOH.

The initial pH is simply the pH of the HCl solution:

+ [H3O ] = [HCl]init = 0.1000 mol/L and pH = -log(0.1000) = 1.00

To calculate the pH after 20.00 mL of NaOH solution has been added:

Initial mol of H O+ = 0.04000 L HCl x 0.1000 mol -3 + 3 = 4.000x10 mol H3O 1 L 0.1000 mol OH- added = 0.02000 L NaOH x = 2.000x10-3 mol OH- 1 L - + The OH ions react with an equal amount of H3O ions, so

+ -3 -3 -3 + H3O remaining = 4.000x10 – 2.000x10 = 2.000x10 mol H3O

The equivalence point occurs when mol of OH- added = initial mol of HCl, so when 40.00 mL of NaOH has been added.

To calculate the pH after 50.00 mL of NaOH solution has been added: 0.1000 mol OH- added = 0.05000 L NaOH x = 5.000x10-3 mol OH- 1 L OH- in excess = 5.000x10-3 – 4.000x10-3 = 1.000x10-3 mol OH-

1.000x10-3 mol [OH-] = = 0.01111 mol/L 0.04000 L + 0.05000 L

pOH = -log(0.01111) = 1.95 pH = 14.00 – 1.95 = 12.05

The pH increases slowly beyond the equivalence point. The curve rises gradually in the buffer region. The weak acid and its conjugate base The pH at the equivalent are both present in solution. point is > 7.00 due to the reaction of the conjugate

base with H2O.

The initial pH is higher Than for the strong acid solution.

weak acid–strong base titration

Initial pH + - [H3O ][A ] + K = [H3O ] = a [HA] + pH = -log[H3O ]

pH before equivalence point

+ [HA] [H3O ] = Ka x or [A-]

[base] pH = pKa + log [acid]

weak acid–strong base titration

pH at the equivalence point

The - image - A (aq) + H2O(l) cannot HA(aq) + OH (aq) The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. - Restart your computer, and then open [OH ] = the ﬁle again. If the red x still mol HA K where [A-] = init and K = w b K Vacid + Vbase a

+ Kw + [H3O ] ≈ and pH = -log[H3O ]

pH beyond the equivalence point mol OH- K [OH-] = excess [H O+] = w 3 [OH-] Vacid + Vbase + pH = -log[H3O ]

17-38 ©2013 McGraw-Hill Ryerson Limited Sample Problem 17.6 Finding the pH During a Weak Acid– Strong Base Titration PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 -5 mol/L propanoic acid (EtCOOH; Ka = 1.3x10 ) after adding the following volumes of 0.1000 mol/L NaOH: (a) 0.00 mL (b) 30.00 mL (c) 40.00 mL (d) 50.00 mL

PLAN: The initial pH must be calculated using the Ka value for the weak acid. We then calculate the number of moles of HPr present initially and the number of moles of OH- added. Once we know the volume of base required to reach the equivalence point we can calculate the pH based on the species present in solution.

SOLUTION: + -3 (a) [H3O ] = = 1.1x10 mol/L pH = -log(1.1x10-3) = 2.96

(b) 30.00 mL of 0.1000 mol/L NaOH has been added.

Initial amount of HPr = 0.04000 L x 0.1000 mol/L = 4.000x10-3 mol HPr Amount of NaOH added = 0.03000 L x 0.1000 mol/L = 3.000x10-3 mol OH-

Each mol of OH- reacts to form 1 mol of Pr-, so - - Concentration (M) HPr(aq) + OH (aq) → Pr (aq) + H2O(l) Initial 0.004000 0.003000 0 - Change −0.003000 -0.003000 +0.003000 - Equilibrium 0.001000 0 0.003000 -

+ [HPr] [H3O ] = Ka x = (1.3x10-5) x 0.001000 = 4.3x10-6 mol/L - [Pr ] 0.003000 pH = -log(4.3x10-6) = 5.37

mol of HAinit. All the OH- added has reacted with HA to form 0.004000 mol of EtCOO-. [EtCOO-] = 0.004000 mol = 0.05000 mol/L 0.04000 L + 0.04000 L K 1.0x10-14 EtCOO- is a weak base, so we calculate w = = 7.7x10-10 . Kb = -5 Ka 1.3x10 K 1.0x10-14 + w -9 [H3O ] ≈ = = 1.6x10 mol/L

pH = -log(1.6x10-9) = 8.80

(d) 50.00 mL of 0.1000 mol/L NaOH has been added.

Amount of OH- added = 0.05000 L x 0.1000 mol/L = 0.005000 mol

- - Excess OH = OH added – HAinit = 0.005000 – 0.004000 = 0.001000 mol

mol OH- 0.001000 mol [OH-] = excess = = 0.01111 mol/L total volume 0.09000 L

-14 + Kw 1x10 -13 [H3O ] = = = 9.0x10 mol/L [OH-] 0.01111

pH = -log(9.0x10-13) = 12.05

The pH decreases The pH at the equivalence gradually in the buffer point is < 7.00 due to the region. The weak base reaction of the conjugate and its conjugate acid acid with H2O. are both present in solution.

Titration of 40.00 mL of 0.1000

mol/L H2SO3 with 0.1000 mol/L NaOH

pKa2 = 7.19

pKa1 = 1.85

Any “insoluble” ionic compound is actually slightly soluble in aqueous solution. We assume that the very small amount of such a compound that dissolves will dissociate completely.

For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions.

2+ - PbF2(s) Pb (aq) + 2F (aq)

[Pb2+][F-]2 Qsp =

Qsp is called the ion-product expression for a slightly soluble ionic compound.

For any slightly soluble compound MpXq, which consists of ions Mn+ and Xz-,

n+ p z- q Qsp = [M ] [X ]

When the solution is saturated, the system is at equilibrium,

and Qsp = Ksp, the solubility product constant.

The Ksp value of a salt indicates how far the dissolution proceeds at equilibrium (saturation).

PROBLEM: Write the ion-product expression at equilibrium for each compound: (a) magnesium carbonate (b) iron(II) hydroxide (c) calcium phosphate PLAN: We write an equation for a saturated solution of each compound, and then write the ion-product expression at

equilibrium, Ksp. Note the sulfide in part (d). SOLUTION: 2+ 2- 2+ 2- (a) MgCO3(s) Mg (aq) + CO3 (aq) Ksp = [Mg ][CO3 ]

2+ - 2+ - 2 (b) Fe(OH)2(s) Fe (aq) + 2OH (aq) Ksp = [Fe ][OH ]

2+ 3- 2+ 3 3- 2 (c) Ca3(PO4)2(s) 3Ca (aq) + 2PO4 (aq) Ksp = [Ca ] [PO4 ]

PROBLEM: (a) Lead(II) sulfate (PbSO4) is a key component in lead- acid car batteries. Its solubility in water at 25°C is -3 4.25x10 g/100 mL solution. What is the Ksp of PbSO4?

(b) When lead(II) fluoride (PbF2) is shaken with pure water at 25°C, the solubility is found to be 0.64 g/L.

Calculate the Ksp of PbF2. PLAN: We write the dissolution equation and the ion-product expression for each compound. This tells us the number of moles of each ion formed. We use the molar mass to convert the solubility of the compound to molar solubility (molarity), then use it to find the molarity of each ion, which we can

substitute into the Ksp expression.

SOLUTION:

2+ 2- 2+ 2- (a) PbSO4(s) Pb (aq) + SO4 (aq) Ksp = [Pb ][SO4 ] Converting from g/mL to mol/L:

-3 4.25x10 g PbSO4 1000 mL x 1 mol PbSO4 -4 x = 1.40x10 mol/L PbSO4 100 mL soln 1 L 303.3 g PbSO4

2+ 2- Each mol of PbSO4 produces 1 mol of Pb and 1 mol of SO4 , so 2+ 2- -4 [Pb ] = [SO4 ] = 1.40x10 mol/L

2+ 2- -4 2 -8 Ksp = [Pb ][SO4 ] = (1.40x10 ) = 1.96x10

2+ - 2+ - 2 (b) PbF2(s) Pb (aq) + F (aq) Ksp = [Pb ][F ] Converting from g/L to mol/L:

0.64 g PbF2 x 1 mol PbF2 -3 = 2.6x10 mol/L PbF2 1 L soln 245.2 g PbF2

2+ - Each mol of PbF2 produces 1 mol of Pb and 2 mol of F , so [Pb2+] = 2.6x10-3 M and [F-] = 2(2.6x10-3) = 5.2x10-3 mol/L

2+ - 2 -3 -3 2 -8 Ksp = [Pb ][F ] = (2.6x10 )(5.2x10 ) = 7.0x10

PROBLEM: Calcium hydroxide (slaked lime) is a major component of

mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a strong, inexpensive base.

Calculate the molar solubility of Ca(OH)2 in water if the -6 Ksp is 6.5x10 .

PLAN: We write the dissolution equation and the expression for Ksp. We know the value of Ksp, so we set up a reaction table that expresses [Ca2+] and [OH-] in terms of S, the molar solubility.

We then substitute these expressions into the Ksp expression and solve for S. SOLUTION:

2+ - 2+ - 2 -6 Ca(OH)2(s) Ca (aq) + 2OH (aq) Ksp = [Ca ][OH ] = 6.5x10

2+ - Concentration(mol/L) Ca(OH)2(s) Ca (aq) + 2OH (aq) Initial - 0 0 Change - +S + 2S Equilibrium - S 2S

2+ - 2 2 3 -6 Ksp = [Ca ][OH ] = (S)(2S) = 4S = 6.5x10

S = = = 1.2x10-2 mol/L

No. of Ions Formula Cation/Anion Ksp Solubility (mol/L) -8 -4 2 MgCO3 1/1 3.5x10 1.9x10 -8 -4 2 PbSO4 1/1 1.6x10 1.3x10 -10 -5 2 BaCrO4 1/1 2.1x10 1.4x10

-6 -2 3 Ca(OH)2 1/2 6.5x10 1.2x10 -6 -3 3 BaF2 1/2 1.5x10 7.2x10 -11 -4 3 CaF2 1/2 3.2x10 2.0x10 -12 -5 3 Ag2CrO4 2/1 2.6x10 8.7x10

The higher the Ksp value, the greater the solubility, as long as we compare compounds that have the same total number of ions in their formulas.

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If Na2CrO4 solution is added to a saturated solution of PbCrO4, it 2- provides the common ion CrO4 , causing the equilibrium to shift to the left. Solubility decreases and solid PbCrO4 precipitates.

Ca(OH)2 in water. What is its solubility in 0.10 mol/L -6 Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10 .

PLAN: The addition of Ca2+, an ion common to both solutions, should

lower the solubility of Ca(OH)2. We write the equation and Ksp expression for the dissolution and set up a reaction table in

terms of S, the molar solubility of Ca(OH)2. We make the 2+ assumption that S is small relative to [Ca ]init because Ksp is low. We can then solve for S and check the assumption.

SOLUTION: 2+ - 2+ - 2 Ca(OH)2(s) Ca (aq) + 2OH (aq) Ksp = [Ca ][OH ]

2+ [Ca ]init = 0.10 mol/L because Ca(NO3)2 is a soluble salt, and dissociates completely in solution.

2+ - Concentration (mol/L) Ca(OH)2(s) Ca (aq) + 2OH (aq) Initial - 0.10 0 Change - +S + 2S Equilibrium - 0.10 + S 2S

2+ - 2 -6 2 2 Ksp = [Ca ][OH ] = 6.5x10 ≈ (0.10)(2S) = (0.10)(4S )

6.5x10-6 4S2 ≈ so S ≈ = 4.0x10-3 mol/L 0.10

-3 Checking the assumption: 4.0x10 mol/L x 100 = 4.0% < 5% 0.10 mol/L

Changes in pH affects the solubility of many slightly soluble ionic compounds. + The addition of H3O will increase the solubility of a salt that contains the anion of a weak acid.

2+ 2- CaCO3(s) Ca (aq) + CO3 (aq)

2- + - CO3 (aq) + H3O (aq) → HCO3 (aq) + H2O(l) - + HCO3 (aq) + H3O (aq) → [H2CO3(aq)] + H2O(l) → CO2(g) + 2H2O(l)

+ The net effect of adding H3O to CaCO3 is the removal 2- of CO3 ions, which causes an equilibrium shift to the right. More CaCO3 will dissolve.

When a carbonate mineral is treated with HCl, bubbles of CO2 form.

PROBLEM: Write balanced equations to explain whether addition of H3O + from a strong acid affects the solubility of each ionic compound: (a) lead(II) bromide (b) copper(II) hydroxide (c) iron(II) sulfide

PLAN: We write the balanced dissolution equation for each compound + and note the anion. The anion of a weak acid reacts with H3O , causing an increase in solubility.

SOLUTION: 2+ - (a) PbBr2(s) Pb (aq) + 2Br (aq) - + Br is the anion of HBr, a strong acid, so it does not react with H3O . The addition of strong acid has no effect on its solubility.

2+ - (b) Cu(OH)2(s) Cu (aq) + 2OH (aq) - OH is the anion of H2O, a very weak acid, and is in fact a strong base. It + will react with H3O : - + OH (aq) + H3O (aq) → 2H2O(l)

The addition of strong acid will cause an increase in solubility.

(c) FeS(s) Fe2+(aq) + S2-(aq) S2- is the anion of HS-, a weak acid, and is a strong base. It will react completely with water to form HS- and OH-. Both these ions will react + with added H3O : - + HS (aq) + H3O (aq) → H2S(aq) + H2O(l) - + OH (aq) + H3O (aq) → 2H2O(l)

The addition of strong acid will cause an increase in solubility.

Qsp = Ksp. When two solutions containing the ions of slightly soluble salts are mixed,

If Qsp = Ksp, the solution is saturated and no change will occur.

If Qsp > Ksp, a precipitate will form until the remaining solution is saturated.

If Qsp =< Ksp, no precipitate will form because the solution is unsaturated.

©2013 McGraw-Hill Ryerson Limited 17-61 Sample Problem 17.12 Predicting Whether a Precipitate Will Form PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a

precipitate form when 0.100 L of 0.30 mol/L Ca(NO3)2 is mixed with 0.200 L of 0.060 mol/L NaF?

PLAN: First we need to decide which slightly soluble salt could form,

look up its Ksp value in Appendix C, and write the dissolution equation and Ksp expression. We find the initial ion concentrations from the given volumes and molarities of the

two solutions, calculate the value for Qsp and compare it to Ksp.

SOLUTION: 2+ - + - + - The ions present are Ca , NO3 , Na , and F . All Na and NO3 salts are -11 soluble, so the only possible precipitate is CaF2 (Ksp = 3.2x10 ).

Ca(NO3)2 and NaF are soluble, and dissociate completely in solution.

We need to calculate [Ca2+] and [F-] in the final solution.

Amount (mol) of Ca2+ = 0.030 mol/L Ca2+ x 0.100 L = 0.030 mol Ca2+. 2+ 2+ 0.030 mol Ca 2+ [Ca ]init = = 0.10 mol/L Ca 0.100 L + 0.200 L

Amount (mol) of F- = 0.060 mol/L F- x 0.200 L = 0.012 mol F-.

- - 0.012 mol F - [F ]init = = 0.040 mol/L F 0.100 L + 0.200 L

2+ - 2 2 -4 Qsp = [Ca ]init[F ] init = (0.10)(0.040) = 1.6x10

-11 Since Qsp > Ksp, CaF2 will precipitate until Qsp = 3.2x10 .

©2013 McGraw-Hill Ryerson Limited 17-63 Sample Problem 17.13 Using Molecular Scenes to Predict Whether a Precipitate Will Form PROBLEM: These four scenes represent solutions of silver (gray) and carbonate (black and red) ions above solid silver carbonate. (The solid, other ions, and water are not shown.)

(a) Which scene best represents the solution in equilibrium with the solid? (b) In which, if any, other scene(s) will additional solid silver carbonate form? (c) Explain how, if at all, addition of a small volume of concentrated strong acid affects the [Ag+] in scene 4 and the mass of solid present. ©2013 McGraw-Hill Ryerson Limited 17-64 Sample Problem 17.13 PLAN: We need to determine the ratio of the different types of ion in

each solution. A saturated solution of Ag2CO3 should have + 2- 2Ag ions for every 1 CO3 ion. For (b) we need to compare 2- + Qsp to Ksp. For (c) we recall that CO3 reacts with H3O .

SOLUTION: + 2- First we determine the Ag /CO3 ratios for each scene. Scene 1: 2/4 or 1/2 Scene 2: 3/3 or 1/1 Scene 3: 4/2 or 2/1 Scene 4: 3/4

+ 2- (a) Scene 3 is the only one that has an Ag /CO3 ratio of 2/1, so this scene represents the solution in equilibrium with the solid.

(b) We use the ion count for each solution to determine Qsp for each one. Since Scene 3 is at equilibrium, its Qsp value = Ksp. 2 2 Scene 1: Qsp = (2) (4) = 16 Scene 2: Qsp = (3) (3) = 27 2 2 Scene 3: Qsp = (4) (2) = 32 Scene 4: Qsp = (3) (4) = 36

Scene 4 is the only one that has Qsp > Ksp, so a precipitate forms in this solution.

+ 2- (c) Ag2CO3(s) 2Ag (aq) + CO3 (aq) 2- + CO3 (aq) + 2H3O (aq) → [H2CO3(aq)] + 2H2O(l) → 3H2O(l) + CO2(g)

+ 2- The CO2 leaves as a gas, so adding H3O decreases the [CO3 ] in solution, causing more Ag2CO3 to dissolve. + [Ag ] increases and the mass of Ag2CO3 decreases.

Selective precipitation is used to separate a solution containing a mixture of ions.

A precipitating ion is added to the solution until the Qsp of the more soluble compound is almost equal to its Ksp. The less soluble compound will precipitate in as large a quantity as possible, leaving behind the ion of the more

soluble compound.

PROBLEM: A solution consists of 0.20 mol/L MgCl2 and 0.10 mol/L - CuCl2. Calculate the [OH ] that would separate the metal -10 ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10 ; -20 Ksp of Cu(OH)2 is 2.2x10 . PLAN: Both compounds have 1/2 ratios of cation/anion, so we can

compare their solubilities by comparing their Ksp values. 10 Mg(OH)2 is 10 times more soluble than Cu(OH)2, so Cu(OH)2 will precipitate first. We write the dissolution equations and Ksp expressions. Using the given cation concentrations, we solve for the [OH-] that gives a saturated 2+ solution of Mg(OH)2. Then we calculate the [Cu ] remaining to see if the separation was successful.

SOLUTION:

2+ - 2+ - 2 -10 Mg(OH)2(s) Mg (aq) + 2OH (aq) Ksp = [Mg ][OH ] = 6.3x10

2+ - 2+ - 2 -20 Cu(OH)2(s) Cu (aq) + 2OH (aq) Ksp = [Cu ][OH ] = 2.2x10

[OH-] = = = 5.6x10-5 mol/L

This is the maximum [OH-] that will not precipitate Mg2+ ion.

Calculating the [Cu2+] remaining in solution with this [OH-] K 2.2x10-20 [Cu2+] = sp = = 7.0x10-12 mol/L [OH-]2 (5.6x10-5)2