Question

Determine the pH of each solution.

a) 0.0200 M HClO4

b)0.120 M HClO2 (for HClO2, Ka=1.1×10−2)

c)0.050 M Sr(OH)2

d)0.0856 M KCN (for HCN, Ka=4.9×10−10)

e)0.165 M NH4Cl (for NH3, Kb=1.76×10−5)

Answer 1

Determine the pH of each solution.

a) 0.0200 M HClO4

HClO4 is a strong acid

[H+] = [HClO4]

[HClO4] = 0.0200 = [H+]

pH = -log(H+) = -log(0.02) = 1.69

b)0.120 M HClO2 (for HClO2, Ka=1.1×10−2)

HClO2 is weak acid so

HClO2 <-> H+ + ClO2-

Ka = [H+][ClO2-]/[HClO2]

1.1*10^-2 = (x*x)/(0.12-x)

x = 0.03245

[H+] =x = 0.03245

pH = -log(0.03245 = 1.49

c)0.050 M Sr(OH)2

[OH-] = 2*[Sr(OH)]

[OH-] = 2*0.05 = 0.10

pOH = -log(0.1) = 1

pH = 14-pOH = 14-1 = 13

d)0.0856 M KCN (for HCN, Ka=4.9×10−10)

KCN --> K+ + CN-

CN- forms hydrolysis

CN- + H2O <-> HCN + OH-

Kb = [HCN][OH-]/[CN-]

Kb = Kw/KA = (10^-14)/(4.9*10^-10) = 0.00002040816 = 2.04*10^-5

2.04*10^-5 = x*x/(0.0856-x)

x = 0.001311

[OH-] = 0.001311

pOH = -log(0.001311) =2.88

pH = 14-2.88 = 11.12

e)0.165 M NH4Cl (for NH3, Kb=1.76×10−5)

NH4Cl -->NH4++ Cl-

NH4+ -- > NH3 + +H*

Ka = [NH3][H+]/[NH4+]

Ka = Kw/Kb = (10^-14)/(1.8*10^-5) = 5.55*10^-10

5.55*10^-10 = x*x/(0.165-x)

x = 9.56*10^-6

[H+] = x = 9.56*10^-6

pH = -log(9.56*10^-6)

pH = 5.01