Question

Calculate the pH of the following solutions.

a.) 0.33 M KHSO4

B.)0.25 M HCN

C.) 0.48 M KOH

D.) 0.067 M HI

E.) 0.36 M NH4NO3

F.)0.50 M Ca(NO3)2

G.)0.080 M C6H5NH2

H.) 0.45 M KNO2

Answer 1

a) 0.33 M KHSO4

KHSO4 in water will form

HSO₄^- H⁺(aq) + SO₄^2-(aq)

Ka for HSO₄^- = 1.30 x 10^-2

Ka = x²/0.33

1.30 x 10^-2 = x²/0.33

x = 0.0655 M

H⁺ conc is 0.0655 M

pH = -log H+

pH = - log 0.0655

pH = 1.18

B.)0.25 M HCN

Ka HCN = 6.2 x 10^-10

Ka = x²/0.25

6.2 x 10^-10 = x²/0.25

x = 1.245 x 10^-5

H⁺ conc is 1.245 x 10^-5 M

pH = -log H+

pH = - log 1.245 x 10^-5

pH = 4.90

C.) 0.48 M KOH

KOH is a strong base

so OH^- is 0.48 M

pOH = -log OH-

pOH = - log 0.48

pOH = 0.32

pH = 14-pOH

pH = 14-0.32

pH = 13.68

D.) 0.067 M HI

This is a strong acid and completely dissociated in solution so its pH is

pH = -log H+

pH = -log 0.067

pH = 1.17