Question

A student performs a titration of a solution of unknown concentration of propanoic acid (Ka = 1.3 x 10-5) using 0.165 M sodium hydroxide solution. If 30.8 mL of the sodium hydroxide solution are required to titrate 36 mL of the propanoic acid solution to the equivalence point, what is the pH of the original propanoic acid solution?

Answer #1

Balanced chemical equation is:

NaOH + C2H5COOH ---> C2H5COONa + H2O

Here:

M(NaOH)=0.165 M

V(NaOH)=30.8 mL

V(C2H5COOH)=36.0 mL

According to balanced reaction:

1*number of mol of NaOH =1*number of mol of C2H5COOH

1*M(NaOH)*V(NaOH) =1*M(C2H5COOH)*V(C2H5COOH)

1*0.165*30.8 = 1*M(C2H5COOH)*36.0

M(C2H5COOH) = 0.1412 M

C2H5COOH dissociates as:

C2H5COOH -----> H+ + C2H5COO-

0.1412 0 0

0.1412-x x x

Ka = [H+][C2H5COO-]/[C2H5COOH]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.3*10^-5)*0.1412) = 1.355*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.355*10^-3 M

So, [H+] = x = 1.355*10^-3 M

use:

pH = -log [H+]

= -log (1.355*10^-3)

= 2.8681

Answer: 2.87

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