# Question 8d970

Dec 3, 2017

Warning! Long Answer. pH = 4.92

#### Explanation:

This type of problem involves two major steps:

First, figure out what is happening during the reaction and what you have in the solution at the end of the reaction.

Second, calculate the pH of that solution.

Step 1. What's happening?

The base is reacting with the acid and forming a salt.

This is a stoichiometry problem, so we must use moles in our calculations.

I usually set up an ICE table to organize my calculations.

$\textcolor{w h i t e}{m m m m m m l} \text{HA" +color(white)(mm) "OH"^"-" → color(white)(mm)"A"^"-" + "H"_2"O}$
$\text{I/mol":color(white)(mll)"0.009 36"color(white)(ml)"0.005 61} \textcolor{w h i t e}{m l m m} 0$
$\text{C/mol":color(white)(m)"-0.005 61"color(white)(m)"-0.005 61"color(white)(m)"+0.005 61}$
$\text{E/mol":color(white)(ml)"0.003 75"color(white)(mmml)0color(white)(mmml)"0.005 61}$

$\text{Moles of HA" = 0.024 color(red)(cancel(color(black)("L HA"))) × "0.39 mol HA"/(1 color(red)(cancel(color(black)("L HA")))) = "0.009 36 mol HA}$

$\text{Moles of OH"^"-"color(white)(l) "added" = 0.017 color(red)(cancel(color(black)("L OH"^"-"))) × ("0.33 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol OH"^"-")))) = "0.005 61 mol OH"^"-}$

The strong base will react completely with the acid.

Now, we put these numbers into the ICE table.

This tells us that we have a solution containing $\text{0.003 75 mol HA}$ and $\text{0.005 61 mol A"^"-}$.

What's a solution of a weak acid and its conjugate base? A Buffer!!!

Now we move to Step 2.

Step 2. Calculate the pH of the buffer

We have the equilibrium

$\textcolor{w h i t e}{m m m m m m l} \text{HA" + "H"_2"O" ⇌ color(white)(mm)"A"^"-"color(white)(m)+ "H"_2"O}$
$\text{E/mol":color(white)(ml)"0.003 75"color(white)(mmmmm)"0.005 61}$

We use the Henderson-Hasselbalch equation:

"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))

"p"K_text(a) = "-log"K_text(a) = "-log"(1.8 × 10^"-5") = 4.74

So,

"pH" = 4.74 + log("0.005 61"/"0.003 75") = 4.74 + log1.50 = 4.74 + 0.17 = 4.92#

Did you notice that I used moles instead of molarities in the above calculation?

I can do that because, if the two substances are in the same solution, $\textcolor{b l u e}{\text{the ratio}}$
$\textcolor{b l u e}{\text{of the molarities is the same as the ratio of the moles}}$.

If you were plotting the titration curve, you would be at the position of the red dot in the graph. 