# Why, when an ammonia solution is titrated with a strong acid, the pH remains relatively constant up to the addition of a stoichiometric quantity of the acid?

Mar 26, 2017

Because you have described formation of a buffer solution.........

#### Explanation:

A $\text{buffer solution}$ is a mixture of a weak acid and its conjugate base in APPRECIABLE quantities, and buffers act to resist GROSS changes in $p H$.

This is seen from the well-known buffer equation, the which we may represent as:

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

Where $\left[{A}^{-}\right]$ is the conjugate base. And $\left[H A\right]$ is its conjugate acid.

In this ammoniacal scenario, the free base is the uncharged species, the ammonia, and the ammonium acid is charged:

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

Many students have difficulties with the logarithmic function. I do not suggest that you do. But I can encourage you and others not to be intimidated when you see a $\log$ function. When I write ${\log}_{a} b = c$, I ask to what power I raise the base $a$ to get $c$. Here, ${a}^{c} = b$. And thus ${\log}_{10} 10 = 1 ,$ ${\log}_{10} 100 = 2 ,$${\log}_{10} {10}^{- 1} = - 1$. And ${\log}_{10} 1 = 0$.

Back in the day, before the advent of electronic calculators, students (and scientists, and engineers, statisticians, and economists) would routinely use log tables to do calculations that a calculator would these days render trivial.