# Question 001e7

Oct 13, 2017

$\text{pH} = 9.17$

#### Explanation:

The thing to remember here is that you can calculate the $\text{pH}$ of a weak base-conjugate acid buffer by using this variation of the Henderson - Hasselbalch equation

"pH" = 14 - underbrace(["p"K_b + log( (["conjugate acid"])/(["weak base"]))])_ (color(blue)(= "pOH"))

In your case, the weak base is ammonia, which has

$\text{p} {K}_{b} = 4.75 \to$ source

Now, notice that your solution contains more conjugate acid--the ammonium cation, ${\text{NH}}_{4}^{+}$--than weak base.

Right from the start, this tells you that the $\text{pOH}$ of the solution will be higher than the $\text{p} {K}_{b}$ of the weak base, which implies that the $\text{pH}$ of the solution will be lower than what you'd get for a buffer that contains equal amounts of ammonia and ammonium cations.

So even without doing any calculations, you can say that

$\text{pH" " " < " " 14 - "p} {K}_{b}$

To find the $\text{pH}$ of the solution, simply plug in your values into the Henderson - Hasselbalch equation

"pH" = 14 - [4.75 + log ( (0.36 color(red)(cancel(color(black)("M"))))/(0.30color(red)(cancel(color(black)("M")))))]#

$\text{pH} = 14 - 4.83$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 9.17}}}$

The answer is rounded to two decimal places, the number of sig figs you have for the concentrations of the weak base and of the conjugate acid.

As predicted, you have

$\text{pH" " " < " } 14 - 4.75$

$9.17 \text{ " < " } 9.25$

This shows that you've made the buffer more acidic by adding a higher concentration of the conjugate acid than the concentration of the weak base.