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# calculate the pH of 0.1M HCOOH (ka =105). To what volume 1 litre solution should be diluted to double pH?

Akshay Meena
7 years ago
1.(a)
HCOOH + H2O <==> H3O+ + HCOO-
initial                 0.10                           0             0
change               -x                              x             x
at equilibrium   0.10-x                          x             x
We already know that ka = 10^5
So, Ka = [H3O+][HCOO-] / [HCOOH] = x^2 / (0.1-x) = 1 x 10^5
Solve for x ==>
x^2 = 10^4 - (10^5)*x
x1=0.0999 , x2 = -100000.0999
So, [H+] = 0.099
pH = -log[H+] = -log(0.099) = 1.0

(b)
After doubling the pH = 2
[H+] = 10^-2
And we know that Kais notconcentration dependent.
Ka = [H3O+][HCOO-] / [HCOOH] = (10^-2)^2 / (c-(10^-2)) = 10^5
c = 0.01 = n1/v2
Initially : c1= 0.1 = n/(1000) ===> n = 100
Now,
c = 0.01 = 100 / v ===> v = 10000 ml = 10L

Thanks & Regards
Akshay Meena,