Chemeng Excercise Solution

Exercise 1. 2. 3. 4. 5. 6.

What is the pH of 0.05 M NH4Cl solution? How many grams of NaHCO3 will be used to make a 1.0 L solution that has pH = 9.0? What is the percent ionization of 0.0055 M aqueous HF? (Ka of HF = 6.8 × 10-4) Calculate the pH of a 1.00 M HNO2 Solution. Calculate the Percent dissociation of a 0.0100M Hydrocyanic acid solution, Ka = 6.20 × 10-10. The weak acid hypochlorous acid is formed in bleach solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the value of the Ka of this weak acid.

What is the [H3O+] of a 0.125 M HClO solution? (Ka of HClO = 3.5 × 10-8 8. Calculate the pH of a 2.0 × 10-3 M solution of NaOH. 9. Ammonia is commonly used cleaning agent in households and is a weak base, with a Kb of 1.8 × 10-5. What is the pH of a 1.5 M NH3 solution? 10. Calculate the pH of a 0.45 M NaCN solution. The Ka value for HCN is 6.2 × 10-10. 7.

Problem: What is the pH of 0.05 M NH4Cl solution? NH4+ (aq)

NH3 (aq) + H+(aq)

Ka = 5.6 × 10-10

Equilibrium concentrations: [NH4+] = 0.05 – x, [H+] = [NH3] = x Assume 0.05 – x = 0.05 to simplify the problem. Ka =

[H+] [NH3] [NH4+]

= 5.6 × 10-10 =

(x) (x) 0.05

x = 5.3 × 10-6 = [H+] = [NH3] pH = - log[H+] = - log(5.3 × 10-6) = 5.28

Problem: How many grams of NaHCO3 will be used to make a 1.0 L solution that has pH = 9.0?

Problem: What is the percent ionization of 0.0055 M aqueous HF? (Ka of HF = 6.8 × 10-4)

Problem: Calculate the pH of a 1.00 M HNO2 Solution H+(aq) + NO2-(aq)

HNO2 (aq)

Ka = 4.0 × 10-4

Equilibrium concentrations: [HNO2] = 1.00 – x, [H+] = [NO2-] = x Ka =

[H+] [NO2-] [HNO2]

= 4.0 × 10-4 =

(x) (x) 1.00 - x

Assume 1.00 – x = 1.00 to simplify the problem. x2 1.00

= 4.0 × 10-4

or x2 = 4.0 × 10-4

x = 2.0 × 10-2 = 0.02 M = [H+] = [NO2-] pH = - log[H+] = - log(2.0 × 10-2) = 2.00 – 0.30 = 1.70

Problem: Calculate the Percent dissociation of a 0.0100M Hydrocyanic acid solution, Ka = 6.20 × 10-10. H3O+(aq) + CN- (aq)

HCN(aq) + H2O(l) Initial 0.0100 Change -x Eq. 0.0100 –x Ka =

0 +x x

0 +x x

[H3O+][CN-] (x)(x) = = 6.20 × 10-10 [HCN] (0.0100-x)

Assume 0.0100 - x ≅ 0.0100 Ka =

x2 0.0100

= 6.2 × 10-10

x = 2.49 × 10-6 % dissociation =

2.49 × 10-6 × 100 = 2.49 × 10-2 0.0100

Problem: The weak acid hypochlorous acid is formed in bleach solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the value of the Ka of this weak acid. [H3O+] = 10-pH = 10-4.19 = 6.46 × 10-5 M

Calculating [H3O+] : Concentration (M) HClO(aq) Initial Change Equilibrium

0.12 -x 0.12 -x

+

H2O(l) ----------

H3O+(aq) + ------+x +x

ClO -(aq) ------+x +x

Assumptions: since HClO is a weak acid, we assume 0.12 - x ≅ 0.12 x = [H3O+] = [ClO-] = 6.46 × 10-5 M Ka =

[H3O+] [ClO-] [HClO]

=

(6.46 × 10-5 M) (6.46 × 10-5 M) 0.12 M

= 3.48 × 10-8

Problem: Hypochlorous acid is a weak acid formed in laundry bleach. What is the [H3O+] of a 0.125 M HClO solution? Ka = 3.5 × 10-8

H3O+(aq) + ClO-(aq)

HClO(aq) + H2O(l) Ka = Concentration (M) Initial Change Equilibrium

[H3O+] [ClO-] = 3.5 × 10-8 [HClO] HClO

H2 O

0.125 -x 0.125 - x

----------

H3O+

ClO-

0 +x x

0 +x x

assume 0.125 - x = 0.125 Ka =

(x)(x) = 3.5 × 10-8 0.125-x

x = 0.661 × 10-4

Problem: Calculate the pH of a 2.0 × 10-3 M solution of NaOH.

Since NaOH is a strong base, it will dissociate 100% in water. NaOH(aq)

Na+(aq) + OH-(aq)

Since [NaOH] = 2.0 × 10-3 M , [OH-] = 2.0 × 10-3 M The concentration of [H+] can be calculated from Kw: [H+] =

Kw = [OH-]

1.0 × 10-14 2.0 × 10-3

= 5.0 × 10-12 M

pH = - log [H+] = - log( 5.0 × 10-12) =12.00 – 0.70 = 11.30

Problem: Ammonia is commonly used cleaning agent in households and is a weak base, with a Kb of 1.8 × 10-5. What is the pH of a 1.5 M NH3 solution? NH4+(aq) + OH-(aq)

NH3 (aq) + H2O(l) Kb = Concentration (M) Initial Change Equilibrium

[NH4+] [OH-] [NH3]

NH3

H2O

NH4+

1.5 -x 1.5 - x

----------

0 +x x

making the assumption: since Kb is small:

Kb =

[NH4+] [OH-]

=

[NH3]

(x)(x) 1.5

= 1.8 × 10-5

Calculating pH: Kw [OH-]

=

1.0 × 10-14 = 1.92 × 10-12 5.20 × 10-3

pH = -log[H3O+] = - log (1.92 × 10-12) = 12.000 - 0.283 pH = 11.72

0 +x x

1.5 M - x = 1.5 M

x = 5.20 × 10-3 = [OH-] = [NH4+]

[H3O+] =

OH-

Problem: Calculate the pH of a 0.75 M NaCN solution. The Ka value for HCN is 6.2 × 10-10. CN-(aq) + H2O(l)

Kb = Kb =

The value of Kb can be calculated from Kw and the Ka value for HCN.

[HCN][OH-] [CN-] Kw Ka (for HCN)

1.0 × 10-14 6.2 × 10-10

=

Concentration (M) Initial Change Equilibrium

Kb = 1.61 × 10-5 =

HCN(aq) + OH-(aq)

CN0.75 -x 0.75 - x

[HCN][OH-] [CN-]

= 1.61 × 10-5 HCN

OH-

0 +x x

0 +x x

=

(x)(x) 0.75 - x

x = [OH-] = 3.47 × 10-3 M pOH = -log[OH-] = 3 – 0.43 = 2.46 pH = 14.00 – 2.46 =11.54

~ =

x2 0.75