# When NH_4Cl mixed with NH_3 what happens?

Mar 24, 2016

You have mixed a weak acid with its conjugate base, and have thus prepared a BUFFER solution.

#### Explanation:

$N {H}_{4}^{+} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s N {H}_{3} \left(a q\right) + {H}_{3} {O}^{+}$

Now ${K}_{a}$ $=$ $\frac{\left[N {H}_{3}\right] \left[{H}_{3} {O}^{+}\right]}{\left[N {H}_{4}^{+}\right]}$.

We can take ${\log}_{10}$ OF BOTH SIDES to give:

${\log}_{10} {K}_{a}$ $=$ ${\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $+$ ${\log}_{10} \left(\frac{\left[N {H}_{3}\right]}{\left[N {H}_{4}^{+}\right]}\right)$.

On rearrangement:

$- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $- {\log}_{10} {K}_{a}$ $+$ ${\log}_{10} \left(\frac{\left[N {H}_{3}\right]}{\left[N {H}_{4}^{+}\right]}\right)$

But by definition, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $p H$, and, $- {\log}_{10} {K}_{a}$ $=$ $p {K}_{a}$.

So,
$p H$ $=$ $p {K}_{a}$ $+$ ${\log}_{10} \left(\frac{\left[N {H}_{3}\right]}{\left[N {H}_{4}^{+}\right]}\right)$

So given concentrations of ammonia and ammonium chloride, the $p H$ of the SOLUTION should remain TOLERABLY CLOSE to the $p {K}_{a}$, which from memory is $4.76$? When $\left[N {H}_{3}\right]$ $=$ $\left[N {H}_{4}^{+}\right]$ then $p H$ $=$ $p {K}_{a}$, because ${\log}_{10} 1$ $=$ $0$.