Question

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?

Answer 1

`"pH" = 9.56`

Explanation:

The first thing to do here is use the base dissociation constant, `K_b`, to determine the `pK_b` of the weak base.

You should know that

`color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_b = - log(K_b))color(white)(a/a)|)))`

Plug in the given `K_b` to get

`pK_b = -log(6.4 * 10^(-5)) = 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 `pK_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 `pK_b`.

The presence of more conjugate acid implies that the solution is more acidic, i.e. the pH is lower than what corresponds to `"pOH" = pK_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")))))`

`"pOH" = 4.19 + log(0.520/0.290) = 4.44`

Since an aqueous solution at room temperature has

`color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))`

you can say that the pH of the buffer is

`"pH" = 14 - "pOH"`

`"pH" = 14 - 4.44 = color(green)(|bar(ul(color(white)(a/a)9.56color(white)(a/a)|)))`