7. Each substance is tested with both blue and red litmus paper. The substance tested is the manipulated variable and the colour change of the litmus is the responding variable.

(1)

7. Each substance is tested with both blue and red litmus paper. The substance tested is the manipulated variable and the colour change of the litmus is the responding variable.

Temperature and concentration are controlled variables.

8. Ɣ Solutions of equal concentration of several bases are prepared and the pH is measured for each solution. The manipulated variable is the base, and the responding variable is the pH. The controlled variables are temperature and concentration.

Ɣ The electrical conductivities of aqueous solutions of equal concentration are measured.

The manipulated variable is the base, the responding variable is the electrical conductivity, and the controlled variables are temperature and concentration.

9. Ammonia and hydronium ions are both pyramidal in shape due to having a lone pair of electrons and three attached hydrogen atoms. However, a hydronium ion is positively charged overall and will repel protons, while an ammonia molecule is electrically neutral overall, and the lone pair will attract a proton.

10. (a)

₂ ₃^2- ^>50% ₂ ^- ₃^-

acid base conjugate base conjugate acid

CH OHCOOH(aq) + CO (aq) m o CH OHCOO (aq) + HCO (aq)

(b) Since glycolic acid solution is 3.9% ionized compared to the 1.3% ionization of acetic acid, glycolic acid would be a “stronger” weak acid. Since the compared ionizations are 3.9% and 1.3%, it is likely that the only difference a consumer might notice is the amount of time that the acid takes to react, and, possibly, how vigorous any signs of the reaction are (such as bubbles produced) etc.

(c) Glycolic acid, which is 3.9% ionized in water at standard ambient temperature and pressure (SATP), would react more quickly than acetic acid, which is 1.3% ionized in water. The glycolic acid, due to its greater ionization in water, produces a larger

hydronium ion concentration and a lower pH. Since the concentration of hydronium ion is larger, the rate of reaction with the rust is faster.

(d) With equally concentrated glycolic acid and acetic acid solutions, they would react with the same amount of rust, because the same chemical amount of acid is present in both solutions. The glycolic acid would react slightly more quickly.

11. (a) CaCO

₃

(s) ĺ CaO(s) + CO

2

(g) (b) CaO(s) + H

₂

O(l) ĺ Ca(OH)

2

(s)

(c) O (aq) + H O(l)

^2– ₂

o OH (aq) + OH (aq)

^– ^–

16.3 ACID–BASE STRENGTH AND THE EQUILIBRIUM LAW

Practice (Page 743)

1. (a) Acid strength depends on the extent to which an acid ionizes in (reacts with) water.

(b) The differing pH, rate of reaction, and conductivity of acids of equal concentration is evidence for differing acid strength.

(c) The strength of an acid refers to how easily a proton can be removed from it. The

stronger the acid, the easier it is to remove a proton from it. The concentration of the acid refers to how much acid is present in the solution relative to the quantity of water. A higher concentration of acid has a larger quantity of acid per unit volume of the solution.

(d) The stronger of two acids does not necessarily make a more acidic solution. Every acid,

when placed into water, reacts with the water to form a conjugate base and a hydronium

ion. A stronger acid will produce more hydronium ions than a weaker acid if they are at

the same initial concentration. It is possible to have a weaker acid produce more

hydronium ions in total than a stronger acid, if the weaker acid is sufficiently more

concentrated.

(2)

2. (a) The quadratic formula is not required, because the acid is a strong acid and the reaction with water is quantitative. The equilibrium hydronium ion concentration is assumed to be numerically equal to the initial acid concentration (assuming 100% ionization).

> @

+

3 equilibrium initial

[H O (aq)] = HBr(aq) 0.20 mol/L

(b) The quadratic formula would be required, because the acid is a weak acid and

4 a

[HF(aq)] 0.20 6.3 10 317

K u

, which is less than 1000.

(c) The quadratic formula is not required, because the acid is a weak acid and

3

5 a

[CH COOH(aq)] 0.20

11 111 1.8 10

K u

, which is greater than 1000.

+ -

-5 3 3

a

3

[H O (aq)][CH COO (aq)]

= 1.8 10 mol/L =

[CH COOH(aq)]

K u

At equilibrium: Let = [H O (aq)] = [CH COO (aq)] x

₃ ⁺ ₃ ^-

Then [CH COOH(aq)] = (0.20 - ) = 0.20

₃

x

(Optional) ICE Table for CH

3

COOH(aq) + H

2

O(l) p H

3

O

⁺

(aq) + CH

3

COO

^í

(aq)

Concentration

[CH

3

COOH(aq) (mol/L)

[H

3

O

⁺

(aq)]

(mol/L)

[CH

3

COO

^-

(aq)]

(mol/L)

Initial 0.20 0 0 Change - x + x + x Equilibrium (0.20 - ) = 0.20 x x x

2

1.8 10 =

-5

0.20 u x

-5 -3

1.8 10 0.20 = 1.9 10

x u u u

+ -3

[H O (aq)] = 1.9 10 mol/L

3

u

According to the equilibrium law, the hydronium ion amount concentration of 0.20 mol/L ethanoic acid is 1.9 10 mol/L u

^-3

.

(d) No quadratic formula is required, because the acid is a strong acid.

> @

+ –3

3 equilibrium 3 initial

[H O (aq)] = HNO (aq) 2.3 10 mol/L u

+ –3

pH = –log H O (aq) = –log (2.3 10 ) = 2.64 ª ¬

3

º ¼ u

(e) The quadratic formula would be required, because the acid is a weak acid and

3 2

3 a

[HNO (aq)] 2.3 10 5.6 10 0.41 K

u

u , which is less than 1000.

(f) The quadratic formula is not required, because the acid is a weak acid and

3 2

8 a

[H S(aq)] 2.3 10

25 843 8.9 10

K

u

u , which is greater than 1000.

+ -

-8 3

a

2

[H O (aq)][HS (aq)]

= 8.9 10 mol/L =

[H S(aq)]

K u

At equilibrium: Let = [H O (aq)] = [HS (aq)] x

₃ ⁺ ^-

Then [H S(aq)] = (2.3 10 - ) = 2.3 10 (using the assumption)

₂

u

³

x u

³

(3)

(Optional) ICE Table for H

2

S(aq) + H

2

O(l) p H

3

O

⁺

(aq) + HS

^í

(aq)

Concentration

[H

2

S(aq)]

(mol/L)

[H

3

O

⁺

(aq)]

(mol/L)

HS

^-

(aq) (mol/L)

Initial 2.3 ×10 0 0

^-3

Change - x + x + x Equilibrium (2.3×10 - ) = 2.3 ×10

^-3

x

^-3

x x

2 -8

8.9 10 =

3

2.3 10 x u

u

-8 3 -5

8.9 10 2.3 10 = 1.4 10

x u u u

u

+ -5

[H O (aq)] = 1.4 10 mol/L

3

u

+ –5

pH = –log H O (aq) = –log (1.4 10 ) = 4.84 ª ¬

3

º ¼ u

According to the equilibrium law, the hydronium ion amount concentration of 2.3 mmol/L hydrosulfuric acid is 1.4 10 mol/L u

^-5

, and the pH of the solution is 4.84.

3. The acid solutions for which a prediction could be calculated, ranked in order of decreasing acidity, are:

HBr(aq), HNO

₃

(aq), CH

₃

COOH(aq), and H

₂

S(aq). This is based on a comparison of the solution pH values, which are calculated as follows:

HBr(aq) pH = -log [H O (aq)] = -log(0.020) = 1.70

₃ ⁺

CH

₃

COOH (aq) pH = -log [H O (aq)] = -log(1.9 10 ) = 2.72

₃ ⁺

u

^–3

HNO

3

(aq) pH = -log [H O (aq)] = -log(2.3 10 ) = 2.64

₃ ⁺

u

^-3

H

₂

S (aq) pH = -log [H O (aq)] = -log(1.4 10 ) = 4.84

₃ ⁺

u

^-5

4. The acids, ranked in order of decreasing acid strength, are: HBr(aq), HNO

₃

(aq), HNO

₂

(aq), HF(aq), CH

₃

COOH(aq), and H

₂

S(aq). The two acids that have essentially the same strength, due to the levelling effect with the water solvent, are HBr(aq) and HNO

₃

(aq).

5. (a)

> @

+ -3

H O (aq)

3

1.16 10 mol/L

= 100% =

HA(aq)

equilibrium initial

p ª ¬ º ¼ u

u 0.100 mol/L u 100% = 1.16 % (b) At equilibrium, [C H COOH(aq)] = (0.100 - 0.00116) mol/L = 0.099 mol/L

₂ ₅

– + –3

2 5 3

[C H COO (aq)] = [H O (aq)] = 1.16 10 mol/L u

+ - 3 2

3 2 5 5

2 5

a

[H O (aq)][C H COO (aq)] (1.16 10 mol/L)

= 1.36 10

[C H COOH(aq)] 0.099 mol/L

K u

u

(c) The K value for propanoic acid is constant at constant temperature (for relatively dilute

_a

solutions). If the temperature of the propanoic acid solution is changed, then the value of the equilibrium constant changes.

6. (a) At equilibrium, [CH CHOHCOO (aq)] = [H O (aq)] = 10

₃ ^– ₃ ⁺ ^pH

10

^2.43

0.0037 mol/L At equilibrium, [CH CHOHCOOH(aq)] = (0.10 - 0.0037) mol/L = 0.10 mol/L

₃

+ +

3 3

3

[H O (aq)] [H O (aq)] 0.0037 mol/L

100% 100% =

[HA(aq)] [CH CHOHCOOH(aq)]

p u u

0.10 mol/L u 100% 3.7%

(b)

+ - 2

4

3 3

3 a

[H O (aq)][CH CHOHCOO (aq)] (0.0037 mol/L)

= 1.4 10

[CH CHOHCOOH(aq)] 0.10 mol/L

K u

(c) In comparing the equilibrium constant for propanoic acid ( K

_a

= 1.35 10 u

^-5

) and

2-hydroxypropanoic acid ( K

_a

= 1.4 10 u

^-4

), the 2-hydroxypropanoic acid is a stronger

acid; it ionizes about ten times as much. This means the hydroxyl group of the

(4)

2-hydroxypropanoic acid must weaken the attraction for the proton of the carboxylic acid group, thus allowing a base to remove that proton more easily.

7. (a) HF(aq) + H O(l) p

₂

H O (aq) + F (aq)

₃ ⁺ ^-

+ -

a 3

[H O (aq)][F (aq)]

= [HF(aq)]

K

(b) The quadratic formula is not required, as the acid is a weak acid and

a 4

[HF(aq)] 2.0 6.3 10 3175

K u

which is greater than 1000.

+ -

4 3

a

[H O (aq)][F (aq)]

6.3 10 =

[HF(aq)]

K u

At equilibrium: Let = [H O (aq)] = [F (aq)] x

₃ ⁺ ^-

Then [HF(aq)] = (2.0 - ) = 2.0 (using the assumption) x

(Optional) ICE Table for HF(aq) + H O(l) p

₂

H O (aq) + F (aq)

₃ ⁺ ^-

Concentration

[HF(aq)]

(mol/L)

[H

3

O

⁺

(aq)]

(mol/L)

[F

^-

(aq)]

(mol/L)

Initial 2.0 0 0 Change - x + x + x Equilibrium (2.0 - ) = 2.0 x x x

2

6.3 10 =

-4

2.0 u x

-4 -2

6.3 10 2.0 = 3.5 10

x u u u

+ -2

[H O (aq)] = 3.5 10 mol/L

3

u

+ -2

pH = -log [H O (aq)] = -log(3.5 10 ) = 1.45

3

u

+ + 2

3 3

[H O (aq)] [H O (aq)] 3.5 10 mol/L

100% 100% =

[HA(aq)] [HF(aq)]

p u

u u

2.0 mol/L u 100% 1.8%

According to the equilibrium law, the amount concentration of hydronium ion is

3.5 u 10

²

mol/L, the amount concentration of hydrofluoric acid is 3.5 u 10

²

mol/L, the pH of the solution is 1.45, and the percent reaction is 1.8%.

8. (a) The quadratic formula is not required, because the acid is a weak acid and

3 4

a 3

[H PO (aq)] 10.0 6.9 10 1449

K u

which is greater than 1000.

+ -

3 3 2 4

3 4

a

[H O (aq)][H PO (aq)]

6.9 10 =

[H PO (aq)]

K u

At equilibrium: Let = [H O (aq)] = [H PO (aq)] x

₃ ⁺ ₂ ₄^-

Then [H PO (aq)] = (10.0 - ) = 10.0 (using the assumption)

₃ ₄

x

(Optional) ICE Table for H PO (aq) + H O(l) p

₃ ₄ ₂

H O (aq) + H PO (aq)

₃ ⁺ ₂ ₄^-

Concentration

[H

3

PO

4

(aq)]

(mol/L)

[H

3

O

⁺

(aq)]

(mol/L)

[H

2

PO

4-

(aq)]

(mol/L)

Initial 10.0 0 0

Change - x + x + x

Equilibrium (10.0 - x ) = 10.0 x x

(5)

2

6.9 10 =

-3

10.0 u x

6.9 10 10.0 = 0.26

-3

x u u

+

[H O (aq)] = 0.26 mol/L

3 +

pH = -log [H O (aq)] = -log(0.26) = 0.58

3

+ +

3 3

3 4

[H O (aq)] [H O (aq)] 0.26 mol/L

100% 100% =

[HA(aq)] [H PO (aq)]

p u u

10.0 mol/L u 100% 2.6%

According to the equilibrium law, the hydronium ion amount concentration of

10.0 mol/L phosphoric acid is 0.26 mol/L, the pH of the solution is 0.58, and the percent reaction is 2.6%.

(b) It is possible that these values are inaccurate, as the calculations take into account only the donation of the first proton from phosphoric acid to water. It is likely that the answers calculated are still close to the actual values, since the K

a

value of H PO (aq) (6.2 u 10

₂ ₄^- ^-8

) is substantially smaller than the K

a

value of H PO (aq) (6.9 u 10

₃ ₄ ^-3

). The phosphoric acid solution is a commercial solution, sold at a very high concentration. K

_a

values are not accurate for very high solution concentrations.

9. (a) At equilibrium, [H O (aq)] = [HC H O (aq)] = 10

₃ ⁺ ₆ ₆ ₆^– ^pH

10

^2.40

0.0040 mol/L ICE Table for H C H O (aq) + H O(l) p

₂ ₆ ₆ ₆ ₂

H O (aq) + HC H O (aq)

₃ ⁺ ₆ ₆ ₆^–

Concentration

[H

2

C

6

H

6

O

6

(aq)]

(mol/L)

[H

3

O

⁺

(aq)]

(mol/L)

[HC

6

H

6

O

6-

(aq)]

(mol/L)

Initial 0.200 0 0 Change - 0.0040 +0.0040 +0.0040 Equilibrium 0.196 0.0040 0.0040

+ - 2

5

3 6 6 6

2 6 6 6

a

[H O (aq)][HC H O (aq)] (0.0040 mol/L)

8.2 10 [H C H O (aq)] 0.196 mol/L

K u

According to the student’s evidence, the K

_a

for ascorbic acid is 8.2 10 u

⁵

.

(b) The student’s result for the K

_a

for ascorbic acid is 8.2 10 u

⁵

, while the value listed in Appendix I is 9.1q10

^í5

. Since the experimental value for the equilibrium constant is lower than the accepted value, and since the K

_a

value increases with increasing

temperature, the temperature of the student’s solution may have been lower than standard temperature (25 qC) when the pH was tested.

Practice (Page 746)

10. Electrical conductivity, pH, and rate of reaction of the solutions are empirical measures that may be used to distinguish strong bases from weak bases.

11. (a) CN (aq) + H O(l) p

^– ₂

HCN(aq) + OH (aq)

^–

–

b –

[HCN(aq)][OH (aq)]

= [CN (aq)]

K

(b) SO (aq) + H O(l) p

₄^2– ₂

HSO (aq) + OH (aq)

₄^– ^–

-

b 4 2

4

[HSO (aq)][OH (aq)]

= [SO (aq)]

K

(6)

12. C H COO (aq) + H O(l) p

₂ ₅ ^– ₂

C H COOH(aq) + OH (aq)

₂ ₅ ^–

At equilibrium, [C

2

H

5

COO

^–

(aq)] = (0.157 – 1.1 u 10

^-5

) mol/L = 0.157 mol/L, and [OH

^–

(aq)] = [C

2

H

5

COO

^–

(aq)] = 1.1 u 10

^-5

mol/L

– -5 2

2 5 -10

b –

2 5

[C H COOH(aq)][OH (aq)] (1.1 10 mol/L)

= = = 7.7 10

[C H COO (aq)] 0.157 mol/L

K u

u 13. C H NH (aq) + H O(l) p

₆ ₅ ₂ ₂

C H NH (aq) + OH (aq)

₆ ₅ ₃⁺ ^–

pOH = 14.00 – pH = 14.00 – 8.81 = 5.19 [OH

^–

(aq)] = 10

^-pOH

= 10

^-5.19

= 6.5 u 10

^-6

mol/L

At equilibrium, [C

₆

H

₅

NH

₂

(aq)] = (0.10 – 6.5 u 10

^-5

) mol/L = 0.10 mol/L, and [OH

^–

(aq)] = [C

₆

H

₅

NH

₃⁺

(aq)] = 1.1 u 10

^-5

mol/L

+ – -6 2

6 5 3 -10 b

6 5 2

[C H NH (aq)][OH (aq)] (6.5 10 mol/L)

= = = 4.2 10

[C H NH (aq)] 0.10 mol/L

K u

u

Section 16.3 Questions (Page 750)

1. Cod(aq) + H O(l) p

₂

HCod (aq) + OH (aq)

⁺ ^–

The quadratic formula is not required, as codeine is a weak base and

6 b

[Cod(aq)] 0.020

11 560 1.73 10

K u

which is greater than 1000

+ –

6 b

[HCod (aq)][OH (aq)]

1.73 10 =

[Cod(aq)]

K u

At equilibrium: Let = [HCod (aq)] = [OH (aq)] x

⁺ ^–

[Cod(aq)] = (0.020 - ) = 0.020 (making the simplifying assumption) x

2

1.73 10 =

–6

0.020 u x

–6 –4

1.73 10 0.020 = 1.9 10

x u u u

[OH

^–

(aq)] = 1.9 u 10

^-4

mol/L

pOH = -log [OH

^–

(aq)] = -log (1.9 u 10

^-4

) = 3.73 pH = 14.00 – pOH = 14.00 – 3.73 = 10.27 2. CN (aq) + H O(l) p

^– ₂

HCN(aq) + OH (aq)

^–

The conjugate acid of cyanide ion is HCN(aq), K

_a

= 6.2 u 10

^-10

For CN

^–

(aq),

–14 –14

–5

b –10

a

1.00 10 1.00 10

= = 1.6 10

6.2 10

K K

u u

The quadratic formula is not required, as cyanide ion is a weak base and

–

5 b

[CN (aq)] 0.18

5000

1.6 10

K u

which is greater than 1000

5 –

b –

[HCN(aq)][OH (aq)]

1.6 10 =

[CN (aq)]

K u

At equilibrium: Let = [HCN(aq)] = [OH (aq)] x

^–

[CN (aq)] = (0.18 - ) = 0.18 (making the simplifying assumption)

–

x

–5 2

1.6 10 =

0.18 u x

(7)

–5 –3

1.6 10 0.18 = 1.7 10

x u u u

[OH

^–

(aq)] = 1.7 u 10

^-3

mol/L

pOH = -log (1.7 u 10

^-3

) = -log (1.7 u 10

^-3

) = 2.77 pH = 14.00 – pOH = 14.00 – 2.77 = 11.23 The pH of the cyanide solution is 11.23.

3. The difficulty in calculating the pH of a saturated 0.018 mol/L solution of acetylsalicylic acid (ASA) is that the simplifying assumption (about the equilibrium concentration of the acid) cannot be used in this case, so the quadratic formula would have to be used to calculate the answer.

6 4 3

4 a

[C H COOCH COOH(aq)] 0.018 3.27 10 55

K u

, which is less than 1000

4. The quadratic formula is not required, as the acid is a weak acid and

3 3 8

10 a

[H BO (aq)] 0.50

8.6 10 5.8 10

K

u

u , which is greater than 1000.

At equilibrium: Let = [H O (aq)] = [H BO (aq)] x

₃ ⁺ ₂ ₃^–

Then [H BO (aq)] = (0.50 - ) = 0.50 (using the assumption)

₃ ₃

x

(Optional) ICE Table for: H BO (aq) + H O(l) p

₃ ₃ ₂

H O (aq) + H BO (aq)

₃ ⁺ ₂ ₃^-

Concentration [H

3

BO

3

(aq)

(mol/L) [H

3

O

⁺

(aq)]

(mol/L) [H

2

BO

3-

(aq)]

(mol/L)

Initial 0.200 0 0 Change - x + x + x

Equilibrium (0.20 - x ) = 0.20 x x

+ –

-10 3 2 3

a

3 3

[H O (aq)][H BO (aq)]

= 5.8 10 mol/L =

[H BO (aq)]

K u

-10 2

5.8 10 = 0.20 u x

-10 –5

5.8 10 0.50 = 1.7 10

x u u u

+ -3

[H O (aq)] = 1.7 10 mol/L

3

u

pH = -log [H

3

O

⁺

(aq)] = -log(1.7 u 10

^-5

) = 4.77

According to the equilibrium law, the pH of the boric acid solution is 4.77.

5. For the initial solution,

6 5

C H OHCOOH

= 1.00 g

n 1 mol

139.14 g u

6 5

= 0.00719 mol 0.00719 mol

[C H OHCOOH(aq)] = = 0.0156 mol/L 0.460 L

At equilibrium,

6 5 2

C H OHCOOH(aq) + H O(l) p H O (aq) + C H OHCOO (aq)

₃ ⁺ ₆ ₅ ^–

– + pH 2.40

6 5 3

C H OHCOO (aq) = H O (aq) = 10

10 0.0040 mol/L

ª º ª º

¬ ¼ ¬ ¼

(An optional ICE table could be included here.)

> C H OHCOOH(aq) = (0.0156 – 0.0040) mol/L = 0.0116 mol/L

6 5

@

+ - 2

3 6 5 -3

a

6 5

[H O (aq)][C H OHCOO (aq)] (0.0040 mol/L)

1.4 10

[C H OHCOOH(aq)] 0.0116 mol/L

K u

(8)

According to the student’s evidence, the K

a

for salicylic acid is 1.4 u 10

^–3

.

6. (The question should refer to the hydrogen ascorbate ion, for which the chemical formula is given. Students should calculate the K

b

of the hydrogen ascorbate ion.)

> @

+ –

6 6 6 6 6 6

–

6 6 6 6 6 6

Assume the dissolved ionic solid dissociates completely:

NaHC H O (s) Na (aq) + HC H O (aq)

and thus, initially, NaHC H O (aq) = HC H O (aq) = 0.15 mol/L o

ª º

¬ ¼

The pH is > 7 (basic), so hydrogen ascorbate ion is reacting (with water) predominantly as a base.

–

6 6 6 2

HC H O (aq) + H O(l) p H C H O (aq) + OH (aq)

₂ ₆ ₆ ₆ ^–

>

² ⁶ ⁶ ⁶

@

^– ^–pOH ^–5.35 ^–6

–

6 6 6

At equilibrium,

pOH = 14.00 - pH = 14.00 - 8.65 = 5.35

H C H O (aq) = OH (aq) = 10 = 10 = 4.5 10 mol/L HC H O (aq) = (0.15 - 0.0000045) mol/L = 0.15 mol/L

ª º u

¬ ¼

ª º

¬ ¼

(An optional ICE table could be included here.)

>

2 6 6 6

@

^– ^–6 ² –10

b –

6 6 6

H C H O (aq) OH (aq) (4.5 10 mol/L)

1.3 10

0.15 mol/L HC H O (aq)

K ª ¬ º ¼ u u

ª º

¬ ¼

According to the K

_W

relationship and the equilibrium law, the value of K

_b

for the hydrogen ascorbate ion is 1.3 u 10

¹⁰

.

7. For a volume of 1 L, assuming the density of the bleach solution to be 1.00 g/mL

(approximately equal to that of pure water), then a litre of this bleach solution would have a mass of 1000 g, and 5.25% of this mass would be sodium hypochlorite.

NaClO

= mass %

NaClO

mass of solution = 5.25% 1000 g = 52.5 g

m u u

+ -

NaClO(aq) o Na (aq) + ClO (aq)

NaClO

= 52.5 g

n 1 mol

74.44 g u

> @

-

= 0.705 mol 0.705 mol

ClO (aq) = NaClO(aq) = = 0.705 mol/L 1.000 L

ª º

¬ ¼

At equilibrium: ClO (aq) + H O(l) p

^– ₂

HClO(aq) + OH (aq)

^–

Let = [OH (aq)] = [HClO(aq)] x

-

(Optional) ICE Table for ClO (aq) + H O(l) p

^– ₂

HClO(aq) + OH (aq)

^–

Concentration [CIO

(aq)]

(mol/L)

[OH

(aq)]

(mol/L)

[HCIO(aq)]

(mol/L)

Initial 0.705 0 0 Change - x + x + x Equilibrium 0.705 - x x x

-14 -7

w

b 8

a

1.00 10

= = 2.5 10 (conjugate acid is HClO(aq)) 4.0 10

K K

K

u u

u

The quadratic formula is not required, as the base is a weak base and

-

8 b

[OCl (aq)] 0.705

2 821 063 4.0 10

K u

, which is greater than 1000.

At equilibrium: [ClO

^–

(aq)] is (0.705 – x) mol/L = 0.750 mol/L (using the assumption).

(9)

-

b -

[HClO(aq)][OH (aq)]

= [ClO (aq)]

K

-7 4

2.5 10 0.705 = 4.2 10

x u u u

- -4

[OH (aq)] = 4.2 10 mol/L u

The amount concentration of hydroxide ion in a 5.25% (by mass) aqueous solution of bleach is 4.2 u 10

^–4

mol/L.

8. CH COOH(aq) + H O(l) p

₃ ₂

H O (aq) + CH COO (aq)

₃ ⁺ ₃ ^–

> @

+ –

3 3

a

3

H O (aq) CH COO (aq)

K CH COOH(aq)

ª º ª º

¬ ¼ ¬ ¼

When water is added to a solution of acetic acid in equilibrium, the concentrations of the acetic acid, hydronium ion, and acetate ion all decrease significantly. The concentration of water does not decrease, however, because the solution is predominantly water. Le

Châtelier’s principle predicts that the equilibrium will shift to the right because there are two entities on the right that have decreased in concentration, and there is only one such entity on the left. Just as would be the case for gases, the shift is toward the side with the most entities of variable concentration, because that shift direction opposes the original change most effectively. This shift to establish the new equilibrium causes the concentration of acetic acid to drop even lower than the original decrease caused by the dilution, while the initial decrease in hydronium ion and acetate ion concentrations is partly counteracted by the equilibrium shift. The end result is that the dilute acid solution has a higher percent ionization than the more concentrated solution, provided that both solutions are at the same temperature.

9. (a) K HPO (s)

₂ ₄

o 2 K (aq) + HPO (aq)

⁺ ₄^2-

2-

HPO (aq) is amphiprotic; the entity reacts with water in aqueous solution both as a base

4

and as an acid. The conjugate acid is H

₂

PO

₄^–

(aq), with a K

_a

value of 6.2 10 u

⁸

. For HPO

₄^2–

(aq),

-14

-7 w

b 8

a

1.00 10

= = 1.6 10

6.2 10 K K

K

u u

u

which is larger than its K

_a

value, of 4.8 10 u

¹³

, so this entity reacts predominantly as a base.

2-

4 2

HPO (aq) + H O(l) p H PO (aq) + OH (aq)

₂ ₄^- ^-

(b) NaH PO (s)

₂ ₄

o Na (aq) + H PO (aq)

⁺ ₂ ₄^-

-

2 4

H PO (aq) is amphiprotic. Its conjugate acid is H

3

PO

4

(aq), with a K

_a

value of 6.9 10 u

3

.

For H

₂

PO

₄^–

(aq),

-14 w -12

b 3

a

1.00 10

= = 1.4 10

6.9 10 K K

K

u u

u

which is smaller than its K

_a

value of 6.2 10 u

⁸

, so it reacts predominantly as an acid.

-

2 4 2

H PO (aq) + H O(l) p H O (aq) + HPO (aq)

₃ ⁺ ₄^2–

(c) Na HC H O (s)

₂ ₆ ₅ ₇

o 2Na (aq) + HC H O (aq)

⁺ ₆ ₅ ₇^2-

2-

6 5 7

HC H O (aq) is amphiprotic. Its conjugate acid is H

₂

C

₆

H

₅

O

₇^–

(aq), with a

5 a

value of 1.7 10

K u

.

For HC

6

H

5

O

72–

(aq),

-14 -10

w

b 5

a

1.00 10

= = 5.9 10

1.7 10 K K

K

u u

u

(10)

which is smaller than its K

_a

value of 6.2 10 u

⁸

, so it reacts predominantly as an acid.

2-

6 5 7 2

HC H O (aq) + H O(l) p H O (aq) + C H O (aq)

₃ ⁺ ₆ ₅ ₇^3-

10. (a) K HPO (s)

₂ ₄

o 2K (aq) + HPO (aq)

⁺ ₄^2-

2-

4 2

HPO (aq) + H O(l) p OH (aq) + H PO (aq) (from question 9)

^- ₂ ₄^-

2– -7

b

for HPO (aq) is 1.6 10

4

K u (from question 9)

The quadratic formula is not required, as hydrogen phosphate ion is a weak base and

- 2

4

7 b

[HPO (aq)] 5.0 10

310 000 1.6 10

K

u

u , which is greater than 1000.

At equilibrium: Let = [OH (aq)] = [H PO (aq)] x

^- ₂ ₄^-

Then [HPO (aq)] = (0.050 - ) = 0.050 (using the assumption)

₄^2-

x

(Optional) ICE Table for HPO (aq) + H O(l) p

₄^2- ₂

OH (aq) + H PO (aq)

^- ₂ ₄^-

Concentration [HPO

42-

(aq)]

(mol/L) [OH

^-

(aq)]

(mol/L) [H

2

PO

4-

(aq)]

(mol/L)

Initial 0.050 0 0 Change - x + x + x Equilibrium (0.050 - x ) = 0.050 x x

- -

2 4

b 2-

4

[OH (aq)][H PO (aq)]

= [HPO (aq)]

K

2

1.6 10

-7

0.050 u x

-7 5

1.6 10 0.050 = 9.0 10

x u u u

- -5

[OH (aq)] = 9.0 10 mol/L u .

- -5

pOH = -log [OH (aq)] = -log(9.0 10 ) = 4.05 u pH = 14.00 pOH = 14.00 - 4.05 = 9.95

According to the equilibrium law, the pH of the solution is 9.95.

(b) NaH PO (s)

₂ ₄

o Na (aq) + H PO (aq)

⁺ ₂ ₄^-

-

2 4 2

H PO (aq) + H O(l) p H O (aq) + HPO (aq) (from question 9)

₃ ⁺ ₄^2-

The quadratic formula is not required, as dihydrogen phosphate ion is a weak acid and

- 2

2 4

8 a

[H PO (aq)] 5.0 10

806 451 6.2 10

K

u

u , which is greater than 1000.

2- +

8 4 3

a -

2 4

[HPO (aq)][H O (aq)]

6.2 10

[H PO (aq)]

K u

At equilibrium: Let = [H O (aq)] = [HPO (aq)] x

₃ ⁺ ₄^2-

Then [H PO (aq)] = (0.050 - ) = 0.050 (using the assumption)

₂ ₄^-

x

(Optional) ICE Table for H PO (aq) + H O(l) p

₂ ₄^- ₂

H O (aq) + HPO (aq)

₃ ⁺ ₄^2-

Concentration

H

2

PO

4-

(aq)]

(mol/L)

[H

3

O

⁺

(aq)]

(mol/L)

[HPO

42-

(aq)]

(mol/L)

Initial 0.050 0 0 Change - x + x + x

Equilibrium (0.050 - x) = 0.050 x x

2

6.2 10 =

-8

0.050 u x

(11)

-8 -5

6.2 10 0.050 = 5.6 10

x u u u

+ -5

[H O (aq)] = 5.6 10 mol/L

3

u

+ -5

pH = -log [H O (aq)] = -log 5.6 10 mol/L = 4.25

3

u

According to the equilibrium law, the pH of the solution is 4.25.

(c) Na HC H O (s)

₂ ₆ ₅ ₇

o 2Na (aq) + HC H O (aq)

⁺ ₆ ₅ ₇^2-

2-

6 5 7 2

HC H O (aq) + H O(l) p H O (aq) + C H O (aq) (from question 9)

₃ ⁺ ₆ ₅ ₇^3-

The quadratic formula is not required, as hydrogen citrate ion is a weak acid and

2- 2

6 5 7

7 a

[HC H O (aq)] 5.0 10

125 000 4.0 10

K

u

u , which is greater than 1000.

3- +

7 6 5 7 3

a 2-

6 5 7

[C H O (aq)][H O (aq)]

4.0 10

[HC H O (aq)]

K u

At equilibrium: Let = [H O (aq)] = [C H O (aq)] x

₃ ⁺ ₆ ₅ ₇^3-

Then [HC H O (aq)] = (0.050 - ) = 0.050 (using the assumption)

₆ ₅ ₇^2-

x

(Optional) ICE Table for HC H O (aq) + H O(l) p

₆ ₅ ₇^2- ₂

H O (aq) + C H O (aq)

₃ ⁺ ₆ ₅ ₇^3-

Concentration [HC

6

H

5

O

7-

(aq)]

(mol/L) [H

3

O

⁺

(aq)]

(mol/L) [C

6

H

5

O

72-

(aq)]

(mol/L)

Initial 0.050 0 0 Change - x + x + x Equilibrium (0.050 - x ) = 0.050 x x

2

4.0 10 =

-7

0.050 u x

-7 -4

4.0 10 0.050 = 1.4 10

x u u u

+ -4

[H O (aq)] = 1.4 10 mol/L

3

u

+ -4

pH = -log [H O (aq)] = -log 1.4 10 mol/L = 3.85

3

u

According to the equilibrium law, the pH of the solution is 3.85.

16.4 INTERPRETING pH CURVES

Practice (Page 754)

1. (a) A buffering region of a pH curve is a relatively level (slope is close to 0) part of the curve preceding or following the equivalence point.

(b) The buffer zone represents the part of the reaction where the titrant that is added to the sample is either much less or much greater than the volume at the equivalence point.

During this portion of the titration, the pH of the solution changes relatively little.

(c) The region near the equivalence point of a pH curve is the steepest (slope is very large) part of the curve between two buffering regions.

(d) The titrant volume at the equivalence point represents the point of the titration where the chemical amounts of titrant and sample reactants are in the same ratio as the

7. Each substance is tested with both blue and red litmus paper. The substance tested is the manipulated variable and the colour change of the litmus is the responding variable.