# Get K for a weak acid equilibrium when given the ionization percentage

There is some weak acid with a concentration of $0.0284$ M, which is ionized by $6.70$%. What is the value of $K_\mathrm a$?

So we got

$$\ce{HA <=> H+ + A-}$$

The formula for the ionization percentage is $\frac{[\ce{H+}]}{[\ce{HA}]}\cdot100$. So since the value is $6.70$ we can solve and get that

$$[\ce{H+}] = 1.90\cdot10^{-3}$$

So now I got concentration of $\ce{H+}$ ions. However, as far as I understand, to calculate $K_\mathrm a$ I need

$$K_\mathrm a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$

I am missing the value of $[\ce{A-}]$. How do I proceed from here?

• Think about the relationship between $\ce{[H+]}$ and $\ce{[A^{-}]}$. – bon Aug 20 '15 at 19:07

For every mole of $\ce{HA}$ that dissociates, you get equimolar amounts of $\ce{H+}$ and $\ce{A-}$. Therefore:

$$[\ce{H+}] = [\ce{A-}] = 1.9\times 10^{-3}\,\mathrm{M}$$

and

$$[\ce{HA}] = (0.0284\,\mathrm{M}) - (1.9\times 10^{-3}\,\mathrm{M}) = 0.0265\,\mathrm{M}$$

and so

$$K_{\mathrm a} = {(1.9\times 10^{-3})^{2}\over 0.0265} = 1.36\times 10^{-4}$$

where the units of molarity are omitted because they cancel due to the implicit presence of $\ce{H2O}$ in the system (which would otherwise show in the denominator).