Question

In a 0.2000 M solution of a monoprotic weak acid, [H+] = 9.86 × 10–4 M. What is the Ka for this acid?

Answer 1

To answer the question we need the equation to calculate Ka (the acid dissociation constant).

For any acid we have the following dissociation equilibrium:

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

The Ka equation is the following:

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

So to answer the question we need the concentration of the acid, which is 0.2M, of the protones, which is 9.86x10-4M, and the cation [A+].

As this is s monoprotic acid, [H+] is the same as [A+], which is 9.86x10-4M.

Now we calculate:

$Ka=\frac{[9.86x10^{-4}M][{9.86x10^{-4}M]}^{{}{}{}}}{[0.2M]}=4.86x10^{-6}M$

So the Ka for this acid is 4.86x10-6 M.