Homework Answers
Solution-
The reaction can be writeen as
H2A ----> H+ + HA- (Ka1 = 4.38x10^-7)
HA- ---> H+ + A2- ... (Ka2 = 5.37x10^-9)
=>4.38*10^-7 = [H+][HA-]/[H2A] = (x)(x)/(0.327 - x)
x2 = 1.43*10^-7 - 4.38*10^-7x
x2 + 4.38*10^-7x - 1.43*10^-7 = 0
Solving the quadratic equation we get
x = 0.00037 M = H+ = HA- (after 1st ionization)
HA- ---> H+ + A2-
I 0.00037 0.00037 0
C- x +x +x
E 0.00037-x 0.00037+x x
5.37*10^-11 = (0.00037 + x)(x)/(0.00037 - x)
=> 5.37*10^-11(0.00037 - x)= 0.00037x + x^2
=> x^2 +1.9869*10^-14- 5.37*10^-11 x-0.00037x=0
=> x^2 -0.0003699x +1.9869*10^-14=0
x = 0.00036 M = [H+] = [A2-] (second ionization)
The ph is determined by first ionization h+=0.00037
[H+] = 0.00037 M
pH = -log 0.00037
= 3.43
Add Answer to:
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