Question

Determine the pH of each of the following solutions:

0.19 M KCHO2

0.21 M CH3NH3I

0.24 M KI

Answer 1

a) KCHO2

in solution K+ and CHO2-

CHO2- + H2O <--> CH2O + OH-

expect basic ph

Kb = [CH2O][OH]/[CHO2-]

kb = kw/ka = (10^-14) / (1.8*10^-4)

5.55*10^-11= x*x / (0.19-x)

x = 3.24*10^-6 = OH-

pOH = -log( 3.24*10^-6) = 5.48

pH = 14- 5.48= 8.52

b)

CH3NH3I

CH3NH3I --> CH3NH3+ + I-

CH3NH3+ H2O <--> CH3NH3OH + H+

Kb = 4.4x10^-4

Ka = [CH3NH3OH][H+]/[CH3NH3+]

ka = kw/kb = (10^-14)/(4.4*10^-4) = 2.27*10^-11

2.27*10^-11 = x*x /(0.21-x)

x = 2.18*10^-6

pH= -log(2.18*10^-6) = 5.66

c)

KI ---> K+ and I-

this wll not form an equilibrium pH = 7