# If a buffer solution is 0.290 M in a weak base (K_b = 6.4 xx 10^(-5)) and 0.520 M in its conjugate acid, what is the pH?

Mar 22, 2016

$\text{pH} = 9.56$

#### Explanation:

The first thing to do here is use the base dissociation constant, ${K}_{b}$, to determine the $p {K}_{b}$ of the weak base.

You should know that

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{p {K}_{b} = - \log \left({K}_{b}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in the given ${K}_{b}$ to get

$p {K}_{b} = - \log \left(6.4 \cdot {10}^{- 5}\right) = 4.19$

So, your buffer solution contains a weak base and its conjugate acid in comparable amounts. As you know, the pOH of a buffer solution that contains a weak abase and its conjugate acid can be calculated using the Henderson - Hasselbalch equation

color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log( (["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))

Notice that when you have equal amounts of weak base and conjugate acid, the pOH of the solution is equal to $p {K}_{b}$.

In your case, you have more conjugate acid than weak base, so right from the start you should expect the pOH of the solution to be higher than $p {K}_{b}$.

The presence of more conjugate acid implies that the solution is more acidic, i.e. the pH is lower than what corresponds to $\text{pOH} = p {K}_{b}$.

So, plug in your values and calculate the pOH of the solution

"pOH" = pK_b + log( (0.520 color(red)(cancel(color(black)("M"))))/(0.290color(red)(cancel(color(black)("M")))))

$\text{pOH} = 4.19 + \log \left(\frac{0.520}{0.290}\right) = 4.44$

Since an aqueous solution at room temperature has

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH " + " pOH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

you can say that the pH of the buffer is

$\text{pH" = 14 - "pOH}$

$\text{pH} = 14 - 4.44 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 9.56 \textcolor{w h i t e}{\frac{a}{a}} |}}}$