# 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?

##### 1 Answer

#### Answer:

#### Explanation:

The first thing to do here is use the *base dissociation constant*,

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

#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

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

The presence of more conjugate acid implies that the solution is **more acidic**, i.e. the pH is **lower** than what corresponds to

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)|)))#