Question

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution is 2.46 2.46 . Calculate the Ka for the acid.

Answer 1

Answer:

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$.

Explanation:

The pH of the solution = 2.46

$pH=-\log[H^+]$

$2.46=-\log[H^+]$

$[H^+]=0.003467 M$

$HA\rightleftharpoons H^++A^-$

Initially

0.0144         0      0

At equilibrium

(0.0144-x)       x       x

The expression if an dissociation constant is given by :

$K_a=\frac{[A^-][H^+]}{[HA]}$

$K_a=\frac{x\times x}{(0.0144-x)}$

$x=[H^+]=0.003467 M$

$K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}$

$K_a=1.099\times 10^{-3}$

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$.