CHEMISTRY 115 ACID-BASE PROBLEMS

CHEMISTRY 113 ACID-BASE PROBLEMS

For each of the following problems, determine the type of problem and give the set-up(s) that will lead to the solution of the problem.

Calculate the pH of a 0.25 M KCN solution. KCN à K⁺ + CN^- ; K⁺ is a spectator ion.

Hydrolysis of a salt of a weak acid

CN^- +	H₂O =	HCN +	OH^-
0.25 - x		x	x

need K_b = K_w/K_a =1.0 x 10^-14 / 6.2 x 10^-10 =1.613 x 10^-5 = [HCN][OH^-] / [CN^-] =[x][x] / [0.25-x]

approximate x as small compared to 0.25 M

x ~ (1.613 x 10^-5 x 0.25)^1/2 = (4.032 x 10^-6)^1/2 = 2.008 x 10^-3; about 0.8% dissociated, approx. ok

[OH^-] ~ 2.008 x 10^-3; pOH = 2.700; pH = 11.30

Calculate the pH of a 0.50 M NaHCO3 solution. NaHCO₃ à Na⁺ + HCO₃^-; Na⁺is a spectator ion; Acid salt; special case;

[H⁺] ~ (Ka₁ x Ka₂)^1/2 ~ (4.5 x 10^-7 x 4.7 x 10^-11)^1/2 ~ (2.115 x 10^-17)^1/2 ~ 4.600 x 10^-9

pH ~ 8.337 ~ 8.34

Exact solution à [H⁺] ~ ({Ka₁K_w + Ka₁Ka₂[HA^-]} / {Ka₁+ [HA^-]})^1/2

= ({4.5 x 10^-21 + 1.058 x 10^-17} / 5.00 x 10^-1)^1/2 = (1.058 x 10^-17 / 0.500)^1/2=(2.116 x 10^-17)^1/2

[H⁺] = 4.6000 x 10^-9; pH = 8.34

Calculate the pH of a 0.25 M HClO3 solution. strong acid; 2 O's not bonded to H

HClO₃ à H⁺ + ClO₃^- ; [H⁺] = 0.25; pH = 0.60₂

Calculate the pH of a 0.50 M NH4Cl solution. NH₄Cl à NH₄⁺ + Cl^-; Cl^- is a spectator ion.

Hydrolysis; salt of a weak base

NH₄⁺ +	H₂O =	H₃O⁺+	NH₃
0.50 - x		x	x

Need K_a =K_w/K_b =1.0 x 10^-14 /1.74 x 10^-5 =5.75 x 10^-10 = [H₃O⁺][NH₃] /[NH₄⁺] =[x][x] /[0.50-x]

Approximate x as small compared to 0.50,

x ~ (5.75 x 10^-10 x 0.50)^1/2 = (2.874 x 10^-10)^1/2 = 1.70 x 10^-5 ~ [H⁺]; approximation okay

pH ~ 4.77

Calculate the pH of a 0.25 M H2SO3 solution. H₂SO₃ = H⁺ + HSO₃^-; HSO₃^- = H⁺ + SO₃^2-;

Dissociation of a weak diprotic acid; Ka₁ = 1.2 x 10^-2; Ka₂ = 6.6 x 10^-8. Because the Ka's are so different, need do only the first dissociation; the H⁺ formed from the second equilibrium will be repressed by that formed in the first. [H⁺] from second dissociation will approximately = Ka₂ or 6.6 x 10^-8 so it will be negligible.

H₂SO₃ =	H⁺ +	HSO₃^-
0.25 - x	x	x

Ka₁ = 1.2 x 10^-2 = [H⁺][HSO₃^-] /[H₂SO₃] = [x][x] /[0.25 -x]; Ka₁ is so large that an approximation is not likely to be good but it will give a ballpark number that can help show math errors in working the quadratic equation.

x ~ (1.2 x 10^-2 x 0.25)^1/2 = (3.0 x 10^-3)^1/2 = 5.48 x 10^-2 ~ [H⁺]; acid is about 22% dissociated; either successive approximations or quadratic equation will work about equally well.

x ~ (1.2 x 10^-2 x (0.25-0.055))^1/2 = x ~ (1.2 x 10^-2 x 0.195)^1/2 = (2.34 x 10^-3)^1/2 = 4.84 x 10^-2 ~ [H⁺];

x ~ (1.2 x 10^-2 x (0.25-0.048))^1/2 = x ~ (1.2 x 10^-2 x 0.2016)^1/2 =(2.42 x 10^-3)^1/2 =4.92 x 10^-2 ~ [H⁺]

x ~ (1.2 x 10^-2 x (0.25-0.049))^1/2 = x ~ (1.2 x 10^-2 x 0.2008)^1/2 =(2.41 x 10^-3)^1/2 4.91 x 10^-2 ~ [H⁺]

or multiplying out and collecting the actual equation,

x² + 0.012x - 3.0 x 10^-3 = 0; x = {-0.012 +/- [0.012² + 4(3.0 x 10^-3)]^1/2} / 2 = {-.012 + .110} / 2

x = 4.91 x 10^-2 = [H⁺]; pH = 1.309 = 1.31; Ka₂ is small compared to .049 so calculation of second dissociation is not necessary

Calculate the pH of a 0.50 M Ba(OH)2 solution. Ba(OH)₂ = Ba²⁺ + 2 OH^-; soluble hydroxide of alkaline earth metals is a strong base.

[OH^-] = 2 x 0.50 M = 1.00 M; pOH = 0.00; pH = 14.00

Calculate the pH of a solution made by adding 40.0 mL of 0.25 M NaOH to 60.0 mL of 0.50 M HCOOH (formic acid). Na⁺ is a spectator ion.

40.0ml x 0.25M = 10 mmol OH^- ; 60.0mL x 0.50M = 30 mmol HCOOH

HCOOH +	OH^- =	H₂O +	HCOO^-
30 - 10 = 20 mmol	10 - 10 = 0 mmol		0 + 10 = 10 mmol

Weak conjugate acid-base pair; BUFFER; best to use the larger of Ka and Kb

HCOOH =	H⁺ +	HCOO^-
20mmol/100mL - x	x	10mmol/100mL + x

K_a = 1.8 x 10^-4 = [H⁺][HCOO^-]/[HCOOH] = [x][0.10 +x] / [0.20 -x] ~ [x][0.10]/[0.20]

x ~ 1.8 x 10^-4 x 0.20 / 0.10 = 3.6 x 10^-4 ~ [H⁺] ; pH = 3.44; expect acidic

8. Calculate the pH of a solution made by adding 40.0 mL of 0.25 M HCOOH to 60.0 mL of 0.50 M NaOH.

40.0ml x 0.25M = 10 mmol HCOOH ; 60.0mL x 0.50M = 30 mmol OH^- (Na⁺ is a spectator)

HCOOH +	OH^- =	H₂O +	HCOO^-
10 - 10 = 0 mmol	30 - 10 = 20 mmol		0 + 10 = 10 mmol

Strong base + weak conjugate base; expect STRONG BASE to dominate since it will repress equilibrium to form OH^- from hydrolysis.

[OH^-] ~ 20 mmol/100 mL = 0.20 M from strong base

K_b = 1.0 x 10^-14 /1.8 x 10^-4 = 5.56 x 10^-11 = [HCOOH][OH^-] /[HCOO^-] ~ [x][0.20]/[0.10] so [OH^-] from hydrolysis ~ 1 x 10^-10 M <<< 0.20 from strong base

pOH = -log(0.20) = 0.70; pH = 13.30; should be strongly basic and is

9. Calculate the pH of a solution made by adding 40.0 mL of 0.25 M HCOOH to 60.0 mL of 0.50 M HCl.

40.0 mLx0.25M = 10mmol HCOOH; 60.0mLx0.50M = 30mmol H⁺ (Cl^- is a spectator ion)

Expect STRONG ACID to dominate; represses equilibrium of weak acid

[H⁺] ~ 30mmol/100mL ~ 0.30 M from strong acid

K_a = 1.8 x 10^-4 = [H⁺][COO^-] /[HCOOH] ~ [0.30][x] /[0.10]; x ~ 1.8 x 10^-4 x 0.10/0.30

~ 6.0 x 10^-5 from weak acid dissociation <<< 0.30M from strong acid.

pH = -log(0.30) = 0.52; should be strongly acidic and is

10. Calculate the pH of a solution made by adding 40.0 mL of 0.30 M HCOOH to 60.0 mL of 0.20 M NaOH.

60.0ml x 0.20M = 12 mmol OH^- ; 40.0mL x 0.30M = 12 mmol HCOOH

HCOOH +	OH^- =	H₂O +	HCOO^-
12 - 12= 0 mmol	12 - 12 = 0 mmol		0 + 12 = 12 mmol

Weak conjugate base; HYDROLYSIS

HCOO^- +	H₂O =	HCOOH +	OH^-
12mmol/100mL - x		x	x

K_b = 1.0 x 10^-14 /1.8 x 10^-4 = 5.56 x 10^-11 = [HCOOH][OH^-] /[HCOO^-] = [x][x] /[0.12 -x]

~ x² / 0.12; x ~ [5.56 x 10^-11(0.25)]^1/2 = (1.39 x 10^-11)^1/2 = 3.73 x 10^-6 ~ [OH^-]<<<0.12 approx. ok

pOH = 5.429; pH = 8.57; with the small Kb, should be weakly basic and it is.

11. Calculate the pH of a solution made by adding 40.0 mL of 0.30 M HCl to 60.0 mL of 0.20 M NaOH. Cl^- and Na⁺ are spectator ions. 40.0 x 0.30 = 12 mmol H⁺; 60.0 x 0.20 = 12 mmol OH^-

H⁺ +	OH^- =	H₂O
12 -12 = 0 mmol	12 -12 = 0 mmol	0 + 12 = 12 mmol

Product is H₂O; since no other acid-base species, dissociation of water dominates;

[H⁺] = [OH^-] = 1.0 x 10^-7; pH = 7.00

12. Calculate the pH of a solution made by adding 40.0 mL of 0.30 M HCl to 60.0 mL of 0.30 M NaOH. Cl^- and Na⁺ are spectator ions. 40.0 x 0.30 = 12 mmol H⁺; 60.0 x 0.30 = 18 mmol OH^-

H⁺ +	OH^- =	H₂O
12 -12 = 0 mmol	18 -12 = 6 mmol	0 + 12 = 12 mmol

EXCESS STRONG BASE; [OH^-] = 6mmol/100mL = 0.06M; pOH = 1.22; pH = 12.8

Expect to be strongly basic and it is.

13. Calculate the pH of a solution made by adding 25.0 mL of 0.50 M NaOH to 50.0 mL of 0.25 M H3PO4. 12.5 mmol OH^-; 12.5 mmol H₃PO₄; 12.5mmol/75.0mL = 0.1667M

H₃PO₄ +	OH^- à	H₂O +	H₂PO₄^-
12.5 -12.5 = 0 mmol	12.5-12.5 = 0 mmol		0 +12.5 = 12.5 mmol

Special case; acid salt only; [H⁺] ~ (Ka₁ x Ka₂)^1/2 ~ (7.1x10^-3x 6.3x10^-8)^1/2 = (4.47x10^-10)^1/2 =

2.11x 10^-5 ~ [H⁺]; pH ~ 4.67

Exact solution à [H⁺] ~ ({Ka₁K_w + Ka₁Ka₂[HA^-]} / {Ka₁+ [HA^-]})^1/2

= ({7.1 x 10^-17 + 7.455 x 10^-11} / 1.73 x 10^-1)^1/2 = (7.455 x 10^-11 / 0.172)^1/2=(4.29 x 10^-10)^1/2

[H⁺] = 2.07 x 10^-5; pH = 4.68

14. Calculate the pH of a solution made by adding 25.0 mL of 0.25 M NaOH to 50.0 mL of 0.50 M H3PO4. 6.25mmol OH^-; 25 mmol H₃PO₄

H₃PO₄ +	OH^- à	H₂O +	H₂PO₄^-
25 -6.25 = 18.75 mmol	6.25-6.25 = 0 mmol		0 +6.25 = 6.25 mmol

Weak conjugate acid-base pair; BUFFER with Ka₁ that has two species.

H₃PO₄ =	H⁺ +	H₂PO₄^-
18.75mmol/75.0mL - x	x	6.25mmol/75.0mL + x

Ka₁ = [H⁺][H₂PO₄^-] /[H₃PO₄] = 7.1 x 10^-3 = [x][0.08333 + x] /[0.25 -x] ~ [x][0.08333] /[0.25]

x ~ 7.1 x 10^-3 x 0.25/0.08333 = 0.0213 M or about 8.5% ionized; approximation no good.

Using successive approximations

x ~ 7.1 x 10^-3 x (0.25 - 0.0213) /(0.0833 + 0.0213) ~ 7.1 x 10^-3 x 0.229/0.105 = 0.0155M

x ~ 7.1 x 10^-3 x (0.25 - 0.0155) /(0.0833 + 0.0155) ~ 7.1 x 10^-3 x 0.234/0.0988 = 0.0168M

x ~ 7.1 x 10^-3 x (0.25 - 0.0168) /(0.0833 + 0.0168) ~ 7.1 x 10^-3 x 0.233/0.100 = 0.0165M

last two answers are within 5% of each other, so it is okay to quit.

[H⁺] ~ 0.0165; pH ~ 1.78. If not refined, pH of 0.0213 = 1.67, which is answer that Henderson-Hasselbach equation would give since it is an approximation.

Using the quadratic equation, 0.0833x + x² =0.001775 - 0.0071x; x² + 0.009043x - .001775 =0

X ={-0.009043 +/-[(.009043)² + 0.0071]^1/2} /2 ={-0.009043 +/-(0.01528)^1/2}/2 =

{-0.009043+ 0.1236} /2 = 0.03317 /2 = 0.0166; pH = 1.78

Calculate the pH of a solution made by adding 50.0 mL of 0.50 M NaOH to 25.0 mL of 0.25 M H3PO4. 25 mmol OH^-; 6.25 mmol H₃PO₄;

base will react as long as acid exists to react with

H₃PO₄ +	OH^- à	H₂O +	H₂PO₄^-
6.25 -6.25 = 0 mmol	25 -6.25 = 18.75 mmol		0 +6.25 = 6.25 mmol
H₂PO₄^- +	OH^- à	H₂O +	HPO₄^2-
6.25 -6.25 = 0 mmol	18.75-6.25 = 12.5 mmol		0 +6.25 = 6.25 mmol
HPO₄^2- +	OH^- à	H₂O +	PO₄^3-
6.25 -6.25 = 0 mmol	12.5- 6.25 =6.25 mmol		0 +6.25 = 6.25 mmol

Now have a strong and a weak base which cannot react further; strong base will dominate

[OH^-] ~ 6.25 mmol/75.0mL = 0.0833 M; pOH ~ 1.08; pH ~ 12.92

The above pH assumes that the hydrolysis of PO₄^3- equilibrium is suppressed so that it does not contribute to the [OH^-]; because K_b = 1.0 x 10^-14/ 4.5 x 10^-13 = 0.0222 is relatively large, this may not be exactly true.

PO₄^3- +	H₂O =	HPO₄^2- +	OH^-
0.0833 - x		x	0.0833 + x

K_b = 0.02222 = [x][0.08333 + x] / [0.08333 -x]; x ~ 0.02222, which is about 30% so approximation is not valid. Using successive approximations,

x ~ 0.02222 x [0.08333-0.02222] /[0.08333+0.02222] = 0.02222 x 0.06111/0.1056 = 0.01129 M

x ~ 0.02222 x [0.08333-0.01129] /[0.08333+0.01129] = 0.02222 x 0.0705/0.0962 = 0.0163 M

x ~ 0.02222 x [0.08333-0.0163] /[0.08333+0.0163] = 0.02222 x 0.0671/0.0996 = 0.0150 M

x ~ 0.02222 x [0.08333-0.0150] /[0.08333+0.0150] = 0.02222 x 0.0684/0.0983 = 0.0155 M

Within 5% so can quit; [OH^-] = .08333 + 0.0155 = 0.0988M; poH = 1.01; pH = 12.99, which is somewhat more basic than assuming that excess strong base alone determines the pH. Note even so, the original approximation is definitely in the ballpark.

16. Calculate the pH of a solution made by adding 50.0 mL of 0.25 M NaOH to 25.0 mL of 0.25 M H3PO4. 12.5 mmol OH^-; 6.25 mmol H₃PO₄

H₃PO₄ +	OH^- à	H₂O +	H₂PO₄^-
6.25 -6.25 = 0 mmol	12.5 -6.25 = 6.25 mmol		0 +6.25 = 6.25 mmol
H₂PO₄^- +	OH^- à	H₂O +	HPO₄^2-
6.25 -6.25 = 0 mmol	6.25-6.25 = 0 mmol		0 +6.25 = 6.25 mmol

The main acid-base species in solution is HPO₄^2-, an acid salt = special case.

[H⁺] ~ (Ka₁ x Ka₂)^1/2 ~ (6.3x10^-8x 4.5 x10^-13)^1/2 = (2.835x10^-20)^1/2 =

1.68x 10^-10 ~ [H⁺]; pH ~ 9.77

M = 6.25mmol/75.0 mL = 0.083333M

Exact solution à [H⁺] ~ ({Ka₂K_w + Ka₂Ka₃[HA^-]} / {Ka₂+ [HA^-]})^1/2

= ({6.3 x 10^-22 + 2.36 x 10^-21} / 8.33 x 10^-2)^1/2 = (2.992 x 10^-21 / 0.08333)^1/2=(3.59 x 10^-20)^1/2

[H⁺] = 1.89 x 10^-10; pH = 9.72. Approximations were not as valid this time so there is more difference between the actual solution and the approximation for this acid salt.

Calculate the pH of a solution made by adding 75.0 mL of 0.25 M NaOH to 25.0 mL of 0.25 M H3PO4. 18.75 mmol OH^-; 6.25 mmol H₃PO₄

H₃PO₄ +	OH^- à	H₂O +	H₂PO₄^-
6.25 -6.25 = 0 mmol	18.75 -6.25 = 12.5 mmol		0 +6.25 = 6.25 mmol
H₂PO₄^- +	OH^- à	H₂O +	HPO₄^2-
6.25 -6.25 = 0 mmol	12.5-6.25 = 6.25 mmol		0 +6.25 = 6.25 mmol
HPO₄^2- +	OH^- à	H₂O +	PO₄^3-
6.25 -6.25 = 0 mmol	6.25- 6.25 = 0 mmol		0 +6.25 = 6.25 mmol

Final product is the weak conjugate base PO₄^3- which will HYDROLYZE

6.25mmol/100mL = 0.0625 M

PO₄^3- +	H₂O =	HPO₄^2- +	OH^-
0.0625- x		x	x

K_b = 1.0 x 10^-14/ 4.5 x 10^-13 = 0.0222 = [HPO₄^2-][OH^-] /[PO₄^3-] = [x][x]/[0.0625 - x]

K is too large and concentration too small for approximation to be good - but it is a ballpark number; x ~ (0.0222 x 0.0625)^1/2 = 0.0373 M; use quadratic equation to solve.

x² + 0.0222x - 0.001389 = 0;

x = {-0.02222 +/- (0.02222² + 0.00555)^1/2} /2 = {-0.0222 + (6.049 x10^-3)^1/2 } /2 =

{-0.0222 + 0.0777} /2 = 0.05555/2 = 0.0278 = [OH^-]; pOH = 1.56; pH = 12.44

(rather than the pH = 12.57 from the approximation)

18. Calculate the pH of a solution made by adding 60.0 mL of 0.25 M NaOH to 25.0 mL of 0.25 M H3PO4. 15 mmol OH^-; 6.25 mmol H₃PO₄

H₃PO₄ +	OH^- à	H₂O +	H₂PO₄^-
6.25 -6.25 = 0 mmol	15 -6.25 = 8.75 mmol		0 +6.25 = 6.25 mmol
H₂PO₄^- +	OH^- à	H₂O +	HPO₄^2-
6.25 -6.25 = 0 mmol	8.75 -6.25 = 2.5 mmol		0 +6.25 = 6.25 mmol
HPO₄^2- +	OH^- à	H₂O +	PO₄^3-
6.25 - 2.5 =3.75 mmol	2.5 - 2.5 = 0 mmol		0 + 2.5 = 2.5 mmol

Weak conjugate acid-base pair; BUFFER using K_b>>Ka₂;

3.75mmol/85.0 mL = 0.0441M HPO₄^2-; 2.5mmol/85.0mL = 0.0294M PO₄^3-

K_b = 1.0 x 10^-14/ 4.5 x 10^-13 = 0.0222 = [HPO₄^2-][OH^-] /[PO₄^3-] = [x +0.0294][x ]/[0.0441 -x]

x ~ 0.0222 x 0.0441/0.0294 = 0.033M, a ballpark number but large compared to 0.029M; use quadratic equation because dissociation is so large.

0.00098 - 0.02222x = x² + 0.0294 x

x² + 0.0516 x - 0.00098 = 0

x = {-0.0516 +/- ( 0.0516² + .00)^1/2} /2 = {-0.0516 +(0.00659)^1/2} /2 = {-0.0516 + 0.0812} /2

= 0.0295 /2 = 0.0148 = [OH^-]; pOH = 1.83; pH = 12.17