# When a weak acid, for which K_a=6,2xx10^-5 is titrated with a strong base, what is the pH at half-equivalence?

Apr 25, 2017

$p H = 4.21$ at half equivalence...................

#### Explanation:

We use the buffer equation,

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$, which is derived here.

But at half equivalence, $\left[H A\right] = \left[{A}^{-}\right]$ BY DEFINITON, and since ${\log}_{10} \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right) = {\log}_{10} \left(1\right) = 0$, then $p H = p {K}_{a}$

${K}_{a} = 6.2 \times {10}^{-} 5$, and this $p {K}_{a} = - {\log}_{10} \left(6.2 \times {10}^{-} 5\right) = 4.21$.

And thus $p H = 4.21$ at the point of half-equivalence. Note that the concentration of the weak acid is irrelevant.