chemistry:solved problems 9

1.   A solution is 0.50M with respect to HCl and 1.00M with respect to formic acid, HCOOH. Calculate the concentration of

the formate ion, HCOO ^–, in this solution. Is the concentration of HCOO ^– found in this solution greater than, less than,

or equal to that found in a 1.00M solution of formic acid? Explain your reasoning. The K_a for formic acid is equal to

1.8 x 10 ^{– 4}.

HCOOH (aq)   +   H₂O (l) W H₃O⁺ (aq)   +   HCOO ^– (aq)

K_a = ([H₃O⁺ ][HCOO ^– ])/[HCOOH] = 1.8 x 10^-4

[HCOOH]                              [H₃O⁺ ]                   [HCOO ^– ]

Initial                           1.00                                     0.50                             0

Change                        – x +x +x

Equilibrium       1.00 – x 0.50 + x x

Since 100(K_a)  < [HCOOH]_initial (0.018 <  1.00) we can assume that 1.00 – x . 1.00 and 0.50 – x . 0.50

1.8 x 10^-4 = (0.50x)/1.00

x = 3.6 x 10^-4

[HCOO ^– ] = 3.6 x 10^-4 M

The concentration of [HCOO ^– ] present in 1.00 M formic acid is calculated as follows.

x²/1.00  =  1.8 x 10^-4 Þ x² = 1.8 x 10^-4 Þ x = 1.3 x 10^-2

[HCOO ^– ] = 1.3 x 10^-2 M

The concentration of formate ion is less in the first solution than that found in 1.00 M formic acid because          the addition of a strong acid suppresses the dissociation (or deprotonation) of a weak acid.

2.   (a) Calculate the % dissociation of 0.0075 M butanoic acid, CH₃CH₂CH₂COOH, (K_a = 1.5 x 10 ^{– 5}).

CH₃CH₂CH₂COOH(aq)   +   H₂O(l) W H₃O⁺(aq)   +   CH₃CH₂CH₂COO ^– (aq)

K_a = ([H₃O⁺ ][CH₃CH₂CH₂COO ^– ])/[CH₃CH₂CH₂COOH] = 1.5 x 10 ^{– 5}

[CH₃CH₂CH₂COOH]                           [H₃O⁺]                   [CH₃CH₂CH₂COO^– ]

Initial                                      0.0075                                         0                                         0

Change                                       – x + x + x

Equilibrium                  0.0075 – x x x

Since 100(K_a) < 0.0075 (0.0015 < 0.0075) Þ 0.0075 – x . 0.0075

1.5 x 10 ^{– 5} = x²/0.0075 Þ x² = 1.1 x 10 ^{– 7}

x = 3.3 x 10 ^{– 4}

ˆ [H₃O⁺ ] = 3.3 x 10 ^{– 4} M

% dissociation = ([H₃O⁺ ]_eq/[CH₃CH₂CH₂COOH]_initial) x 100%

% dissociation = (3.3 x 10 ^{– 4} M /0.0075M) x 100%

% dissociation = 4.4 %

(b) Calculate the % dissociation of 0.0075 M butanoic acid in a solution containing 0.0025 M sodium butanoate.

[CH₃CH₂CH₂COOH]                           [H₃O⁺]                   [CH₃CH₂CH₂COO^– ]

Initial                                      0.0075                                         0                                 0.0025

Change                                       – x + x + x

Equilibrium                  0.0075 – x x 0.0025 + x

Since 100(K_a) < 0.0025 (0.0015 < 0.0025) Þ 0.0075 – x . 0.0075 and 0.0025 + x . 0.0025

1.5 x 10 ^{– 5} = 0.0025x/0.0075

x = 4.5 x 10 ^{– 5}

ˆ [H₃O⁺ ] = 4.5 x 10 ^{– 5} M

% dissociation = ([H₃O⁺ ]_eq/[CH₃CH₂CH₂COOH]_initial) x 100%

% dissociation = (4.5 x 10 ^{– 5} M /0.0075M) x 100%

% dissociation = 0.60 %

3.   A buffer is prepared by adding 0.15 mol of lactic acid, CH₃CHOHCOOH, and 0.20 mol of sodium lactate,

Na⁺CH₃CHOHCOO ^–, to sufficient water to make 1.0 L of buffer solution. The K_a of lactic acid is 1.4 x 10 ^{– 4}.

pK_a = – logK_a = – log(1.4 x 10^-4) = 3.85

You can use the Henderson-Hasselbalch equation for this problem.

pH = pK_a +  log([lactate ion]/[lactic acid])

(a) Calculate the pH of the buffer.

[lactic acid] = 0.15 mol/1.0L = 0.15 M [lactate ion] = 0.20 mol/1.0 L = 0.20 M

pH = pK_a + log(0.20/0.15) = 3.85 + log(1.3) = 3.85 + 0.12

pH = 3.97

(b) Calculate the pH of the original buffer after 0.050 mol of HCl has been added.

Adding strong acid so [lactic acid] ­ and  [lactate ion]¯.

[lactic acid] = 0.15M +  0.050M =  0.20M [lactate ion] = 0.20M –  0.050M =  0.15M

pH = 3.85 + log(0.15/0.20) = 3.85 + log(0.75) =  3.85 – 0.12

pH = 3.73

(c) Calculate the pH of the original buffer after 0.050 mol of NaOH has been added.

Adding strong base so [lactic acid] ¯ and  [lactate ion] ­.

[lactic acid] = 0.15M –  0.050M =  0.10M [lactate ion] = 0.20M +  0.050M =  0.25M

pH = 3.85 + log(0.25/0.10) = 3.85 + log(2.5) =  3.85 + .40

pH = 4.25

4.   You have to prepare a pH 3.50 buffer, and you have the following 0.10 M solutions available: HCOOH, CH₃COOH,

ClCH₂COOH, NaHCOO, NaCH₃COO, and NaClCH₂COO. Which solutions would you use? For the best system,

calculate the ratio of the concentrations of the buffer components required to prepare the buffer.           

You need to recognize that there are three conjugate acid-base pairs present.

1. HCOOH/NaHCOO                                          (pK_a for HCOOH = 3.74)

2. CH₃COOH/NaCH₃COO                                  (pK_a for CH₃COOH = 4.74)

3. ClCH₂COOH/NaClCH₂COO                          (pK_a for ClCH₂COOH = 2.85)

The effective pH range for each of these buffer systems is given below.

Buffer 1: 2.74 – 4.74

Buffer 2: 3.74 – 5.74

Buffer 3: 1.85 – 3.85

Buffer 2 is ruled out because the desired pH of 3.50 is outside of the effective range of this buffer. Even              though the desired pH falls within the effective range of Buffer 3, Buffer 1 would be the best system since     the pK_a value is nearest to 3.50.

You need to use the Henderson-Hasselbalch equation for this part of the problem.

pH = pK_a +  log([conjugate base]/[weak acid])

log([conjugate base]/[weak acid]) = pH – pK_a

log([conjugate base]/[weak acid]) = 3.50 – 3.74 = – 0.24

[conjugate base]/[weak acid]  = 10 ^{– 0.24}

[conjugate base]/[weak acid]  = 0.58

Therefore the conjugate base-to-weak acid ratio required for this buffer solution is 0.58:1

5.   Label the equivalence point and the half-equivalence point on the titration curve given below. Determine the pK_a and K_a

of the weak acid.

Notice that since the pH at the equivalence point is greater than 7 we are dealing with a weak acid. The              value of the pH at the half-equivalence point is equal to the pK_a value of that weak acid.

pK_a = 4.25

K_a = 10 ^{– pKa} = 10 ^{– 4.25} = 5.6 x 10 ^{– 5}

6.  Write balanced equations and expressions for K_sp for the dissolution of each of the following ionic compounds:

(a) NiCO₃

NiCO₃(s) W Ni ²⁺(aq)   +   CO₃ ^2-(aq)

K_sp = [Ni ²⁺ ][CO₃ ^2- ]

(b) BaF₂

BaF₂(s) W Ba ²⁺(aq)   +   2 F ^–(aq)

K_sp = [Ba ²⁺ ][F ^– ]²

(c) Ag₃PO₄

Ag₃PO₄(s) W 3 Ag ⁺(aq)   +   PO₄ ^3-(aq)

K_sp = [Ag ⁺ ]³[PO₄ ^3-]

7.  Use the K_sp value given in Table C.4 of Appendix II to estimate the solubility of calcium fluoride, CaF₂, (a) in moles per

liter and (b) in grams per 100.0 mL of pure water.

CaF₂(s) W Ca ²⁺(aq)   +   2 F ^–(aq)

K_sp = [Ca ²⁺ ][F ^– ]² = 1.46 x 10 ^{– 10}

(a)      If we let s = [Ca ²⁺ ] = molar solubility, then 2s =  [F ^– ].

K_sp = (s)(2s)² = 4s³ = 1.46 x 10 ^{– 10}

s³ = 3.65 x 10 ^{– 11}

s = 3.32 x 10 ^{– 4}

Thus, the molar solubility of calcium fluoride is estimated to be 3.32 x 10 ^{– 4} M.

(b)      molar mass of CaF₂ = 78.08 g/mol

(3.32 x 10 ^{– 4} mol/L)(78.08 g/mol) = 0.0259 g/L = 0.0259 g/1000mL = 0.00259 g/100.0 mL

8.  Use the K_sp value given in Table C.4 of Appendix II to estimate the molar solubility of BaCO₃ at  25EC in (a) pure water,

and (b) a solution that is 0.25 M in K₂CO₃. Explain why the two values differ.

BaCO_{3 (s)} W Ba ²⁺_(aq) +   CO₃ ^2-_(aq)

K_sp = [Ba ²⁺ ][CO₃ ^2- ] = 2.58 x 10 ^{– 9}

(a) Let s = [Ba ²⁺ ] = [CO₃ ^2- ]

s² = 2.58 x 10 ^{– 9}

s = 5.08 x 10 ^{– 5}

Therefore, the molar solubility of BaCO₃ in pure water is 5.08 x 10 ^{– 5} M.

(b) Potassium carbonate is a strong electrolyte so it is another source of carbonate ions.

K₂CO_{3 (aq)} ® 2 K ⁺ _(aq) +   CO₃ ^2-_(aq)

From the chemical formula it is seen that the concentration of carbonate ions is equal to the concentration        of potassium carbonate.

If we let s = [Ba ²⁺ ], then [CO₃ ^2- ] = 0.25 + s

Since 100(K_sp) < [CO₃ ^2- ]_initial (2.58 x 10 ^{– 7} < 0.25) we can assume that 0.25 + s is essentially 0.25.

K_sp = [Ba ²⁺ ][CO₃ ^2- ] = s x 0.25 = 2.58 x 10 ^{– 9}

s = (2.58 x 10 ^{– 9})/0.25 = 1.0 x 10 ^{– 8}

The molar solubility of BaCO₃ in a solution that is 0.25 M in K₂CO₃ is 1.0 x 10 ^{– 8} M.  The presence of the                 common ion CO₃ ^2- shifts the equilibrium to the left decreasing the             solubility of barium carbonate.

9.  If the concentration of Zn²⁺ in 10.0 mL of water is 1.63 x 10 ^{– 4} M, will zinc hydroxide, Zn(OH)₂, precipitate when 4.0 mg of

NaOH is added? The K_sp value for zinc hydroxide can be found in Table C.4 of Appendix II.

Zn(OH)₂ (s) W Zn ²⁺ (aq)   +   2 OH ^– (aq)

K_sp = [Zn ²⁺ ][OH ^– ]² = 3 x 10 ^{– 17}

We need to first determine [OH ^– ]. With knowledge of [OH ^– ] and [Zn ²⁺ ] we can calculate Q_sp. Comparison         of Q_sp with K_sp allows us to determine whether zinc hydroxide will precipitate under these conditions.

Since sodium hydroxide is a strong base, it is a strong electrolyte.

NaOH (s) ® Na ⁺ (aq)   +   OH ^– (aq)

From the chemical formula it is seen that the concentration of hydroxide ions is equal to the concentration        of sodium hydroxide.

moles OH ^– = moles NaOH = (4.0 mg)(1 g/1000 mg)(1 mol/40.0 g) = 1.0 x 10 ^{– 4} mol

10.0 mL(1 L/1000 mL) =0.010 L

[OH ^– ] = (1.0 x 10 ^{– 4} mol)/0.010 L = 0.010 M

Q_sp = (1.63 x 10 ^{– 4})(0.010)² = 1.6 x 10 ^{– 8}

Zinc hydroxide will precipitate since Q_sp is greater than K_sp (1.6 x 10 ^{– 8} > 3 x 10 ^{– 17}).