Question

Calculate the pH of a 0.1280 M aqueous solution of potassium hydrogen selenide, KHSe.

The first and second ionization constants for H₂Se are 1.3×10^-4 and 1.0×10^-11

pH =

Answer 1

KHSe ---------------- K+ + HSe-

[HSe-]= 0.1280M

HSe- + H2O ---------------------- H2Se + OH-

0.1280                                           0        0

-x                                                 +x      +x

0.1280-x                                     +x        +x

it is the second ionisation process

Ka2= 1.0x10^-11

Ka= 1.0x10^-11

KaxKb= Kw              where Kw=ionic product of water = 1.0x10^-14

Kb= Kw/Ka= 1.0x10^-14/1.0x10^-11= 1.0x10^-3

Kb= 1.0x10^-3

Kb= [H2Se][OH-]/[Hse-]

1.0x10^-3 = x*x/(0.1280-x)

for solving the equation

x=0.0108

[OH-] = 0.0108M

-log[OH-]= -log(0.0108)

POH=1.97

PH+POH=14

PH= 14-POH

PH= 14-1.97

PH=12.03