Question

Be sure to answer all parts.

Answer 1

The dissociation constant is 2.44 mM.

Explanation:

Dissociation constant of any acid or any kind of solutions is the measure of amount dissolved in the solution or separated into different elements. So it is calculated as the ratio of product of concentration of dissociated elements to the concentration of the original compound.

Here the concentration of the acid [HA] = 0.035 M, the pH of the acid is given as 4.67.

$pH=-log[H^{+}]$

$[H^{+}]=e^{-pH} = \frac{1}{e^{pH} } = \frac{1}{e^{4.67} } =\frac{1}{107}$

$[H^{+}]=9.35 \times 10^{-3} M$

Then, $K_{a} =\frac{[H^{+}][A^{-}] }{[HA]} =\frac{(9.35 \times 10^{-3}) ^{2} }{0.035}\\ \\K_{a} = 2497.78 \times 10^{-6}$

Thus, the dissociation constant is 2.44 mM.