# Why does pH = pK when pH = 7, according to the Henderson-Hasselbalch equation?

Nov 11, 2015

In fact $p H$ $=$ $p {K}_{a}$ when $\left[H A\right]$ $=$ $\left[{A}^{-}\right]$; i.e. $p H$ $=$ $p {K}_{a}$ at the point of half-equivalence.

#### Explanation:

We may represent the behaviour of weak acid in water by the following equation:

$H A \left(a q\right) r i g h t \le f t h a r p \infty n s {H}^{+} + {A}^{-}$. And we write the equilibrium reaction as follows,

${K}_{a} = \frac{\left[{H}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$

We take ${\log}_{10}$ of both sides:

${\log}_{10} \left({K}_{a}\right) = {\log}_{10} \left[{H}^{+}\right] + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

On rearrangement,

$- {\log}_{10} \left[{H}^{+}\right] = - {\log}_{10} \left({K}_{a}\right) + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

OR

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

(because by definition $- {\log}_{10} \left[{H}^{+}\right] = p H$, and $- {\log}_{10} \left({K}_{a}\right) = p {K}_{a}$).

I have no doubt you've seen this equation before, but what does it mean? What, precisely does the log function mean? When I write ${\log}_{a} b = c$, it means to get $c$ I have to raise the base (of the log) $a$ to the ${c}^{t h}$ power to get $b$, i.e. ${a}^{c} = b$. Before the advent of electronic calculators, student were routinely issued log books and used to do complex mulitplications and divisions by taking logs and antilogs. Note that ${\log}_{10} 1$ $=$ $0$. So in the equation above, at the point of half-equivalence $\left[{A}^{-}\right] = \left[H A\right]$, and the log of this quotient $= 0$. Thus at half-equivalence, $p H = p {K}_{a}$.

See if you can get your head round this. In the meantime, can you tell me values for ${\log}_{10} 10$, ${\log}_{10} 100$, ${\log}_{10} 1000$, and ${\log}_{10} 1000000$? Use a calculator if you are stumped; why are these values simple whole numbers?