Question

What is the pH of a 0.1M solution of benzoic acid? ( Write answer to the hundredths place) K_A(benzoic acid) = 6.46x10^-5

Answer 1

Lets see the dissociation reaction of benzoic acid

C₆H₅COOH $\rightleftharpoons$ H⁺ + C₆H₅COO^-  

we have

[C₆H₅COOH ] = 0.1 M

Ka = 6.46 $\times$ 10^-5  

Lets see the ICE table for above reaction

..........................C₆H₅COOH....$\rightleftharpoons$.....H⁺ ......+ ....C₆H₅COO^-  

Initial.....................0.1 M.....................0......................0

Change ................-x...........................+x....................+x

At Equilibrium.....0.1-x........................x......................x

The equilibrium constant for above reaction is

Ka = [H⁺][C₆H₅COO^-] / [C₆H₅COOH]

Hence

6.46 $\times$ 10^-5 = (x)(x) /(0.1-x) = x² / (0.1-x)

as x is too small than 0.1 then 0.1-x = 0.1

so

x² / 0.1 = 6.46 $\times$ 10^-5

x² = 0.1 $\times$ 6.46 $\times$ 10^-5

x²   = 6.46 $\times$ 10^-6

by taking square roots on both the side , we get

x = 0.0025

then

[H⁺] = x = 0.0025 M

now we know that

pH = -log[H⁺]

= -log(0.0025)

= 2.60

so our answer is

pH = 2.60