Question

Calculate the [H⁺] of the following mixtures.

a.Dilution and Common Ion problem: 25 mL of 0.10 M HClO₄ and 25 mL of 0.15 M HClO

 

b.Buffer problem: 25 mL of 0.025 M lactic acid (for which K_a = 1.37 × 10^–4) and 25 mL of 0.015 M NaOH

 

c.Dilution and Weak Acid problem: 25 mL of 0.10 M ascorbic acid and 25 mL of water

 

d.Dilution and Conjugate Acid problem: 25.0 mL of 2.50 × 10^-3 M nicotine (for which K_b = 1.05 × 10^-6) and 50.0 mL of 1.25 × 10^-3 M HCl

 

Answer 1

a)

totalV = 25+25 = 50

mmol of HClO4 total = 25*0.1 + 25*0.15 = 6.25

[HClO4] = mmol/V = 6.25/50 = 0.125 M

[H+] = [HClO4] = 0.125 M

b)

mmol of acid = MV = 0.025*25 = 0.625

mmol of base = MV = 0.015*25 = 0.375

after reaction

mmol of acid left = 0.625-0.375 = 0.25

mmol of conjguate formed = 0.375

pH = pKa + log(A-/HA)

pKa = -log(Ka) = -log(1.37*10^-4) = 3.8632

pH = 3.8632+log(0.375/0.25)

pH = 4.03929

[H+] = 10^-pH = 10^-4.03929 = 0.0000913

c)

Vfinal = 25+25 = 50 mL

[Acid] = 0.1 M

Ka = [H+][H2A-]/[[H3A]

10^-3.15 = x*x/(0.1-x)

x = [H+] = 0.008067

d)

mmol of base = MV = (2.5*10^-3)(25) = 0.0625

mmol of acid = MV = (1.25*10^-3)(50) = 0.0625

there is neutralization:

[nicotine acid] = mmol/V = (0.0625)/(25+50) = 0.00083

Let B --> CH3NH2 and BH+ = CH3NH3+ for simplicity

the next equilibrium is formed, the conjugate acid and water

BH+(aq) + H2O(l) <-> B(aq) + H3O+(aq)

The equilibrium is best described by Ka, the acid constant

Ka by definition since it is an base:

Ka = [H3O][B]/[BH+]

get ICE table:

Initially

[H3O+] = 0

[B] = 0

[BH+] = M

the Change

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = -x

in Equilibrium

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

Ka = Kw/Kb = (10^-14)/(1.05*10^-6 = 9.523*10^-9

9.523*10^-9 = x*x/(0.00083-x)

solve for x

x^2 + Ka*x - M*Ka= 0

solve for x with quadratic equation

x = H3O+ =2.81*10^-6

[H+] = 2.81*10^-6 M