Answer to Question #313745 in Organic Chemistry for swarnajeet kumar

Question

Calculate the pH of 0.01 M aqueous solution of sodium formate at 298 K. [Given: Ka (HCOOH) =1.7 × 10−4 at 298K] 

Answer 1

• The dissociation constant Ka= C x²/(1-x), where x = degree of dissociation.
• For a weak acid the [H+]= C x, where C= concentration. In very dilute conditions x <<<1.
• So (1-x)~= 1. So x= √(Ka/C).
• So x= (1.77 x10^-4)/0.001)^0.5 =0.4207.
• Therefore [H+]= 0.001 x 0.4207 =4.207 x10^-4.
• So pH=- log(4.207 x 10^-4) = 3.376