Answer: The pH of the solution is 1.703
Explanation:
We are given:
Concentration of monochloroacetic acid = 0.31 M
The chemical equation for the dissociation of monochloroacetic acid follows:
[tex]HC_2H_2ClO_2\rightleftharpoons H^++C_2H_2ClO_2^-[/tex]
Initial: 0.31
At eqllm: 0.31-x x x
The expression of [tex]K_a[/tex] for above equation follows:
[tex]K_a=\frac{[H^+][C_2H_2ClO_2^-}}{[HC_2H_2ClO_2]}[/tex]
We are given:
[tex]K_a=1.35\times 10^{-3}[/tex]
Putting values in above equation follows:
[tex]1.35\times 10^{-3}=\frac{x\times x}{(0.31-x)}\\\\x=-0.021,0.0198[/tex]
Neglecting the negative value of 'x' because concentration cannot be negative.
To calculate the pH of the solution, we use the equation:
[tex]pH=-\log[H^+][/tex]
[tex]pH=-\log (0.0198)\\\\pH=1.703[/tex]
Hence, the pH of the solution is 1.703
