A weak acid with a Ka of 1.8E-5 is titrated with a strong base. During the titration, 12.5 mL of 0.10 M N(base a)OH is added to 50.0 mL of 0.100 M acetic acid. What is the pH after the addition of the N(base a)OH?

Question

A. 4.26
B. 5.13
C. 7.22
D. 8.59

I have no idea how to do this. Can someone help me.

Answer 1

That a you're calling a base a is a small a to form the symbol for sodium, Na, and NaOH is sodium hydroxide. Let's call acetic acid HAc. Then the titration
millimols HAc = mL x M = 50.0 x 0.100 M = 5
millimols NaOH = 12.5 mL x 0.1 M NaOH = 1.25
.................HAc + NaOH ==> NaAc + H2O
I.................5...............0..............0..........0
add...........................1.25................................
C................-1.25.....-1.25........+1.25.....+1.25
E................3.75.........0...............1.25

Total volume = 50 + 12.5 = 62.5 mL
pH = pKa + log (NaAc)/(HAc).
Note the E line in millimols.
(NaAc) = millimols NaAc/total volume in mL.
(HAc) = millimols HAc/total volume in mL.
pKa for HAc = -log Ka
Substitute and solve for pH.
Post your work if you get stuck.

Answer 2

so for the (NaAC) = 1.25/62.5 mL = 0.02
for the (HAc) = 3.75/62.5 = 0.06

pKa for HAc = -log (1.8E-5) = 4.74

pH = 4.74 + (-0.48) = 4.26.
So the Answer is A right.

Also can you please explain to me how to create the ICE table. I didn't understand that.

Answer 3

Yes, I would go with A. Here is the copied ICE table. It's a little tough to explain chemistry virtually but my best try is below the ICE chart.
millimols HAc = mL x M = 50.0 x 0.100 M = 5
millimols NaOH = 12.5 mL x 0.1 M NaOH = 1.25
.................HAc + NaOH ==> NaAc + H2O
I.................5...............0..............0..........0
add...........................1.25................................
C................-1.25.....-1.25........+1.25.....+1.25
E................3.75.........0...............1.25
The first line is just the reaction between acetic acid (CH3COOH which I'm calling HAc to keep typing and spacing problems to a minimum) and NaOH. Acids + bases ==> salt + water.
Then the ICE stands for I = initial, C = change and E = equilibrium.
In this case we started with 5 millimols HAc and before it has done its thing and before we've added any NaOH, you have zero NaOH, 0 NaAc (the salt), and 0 H2O. Then we add, as a separate line, 1.25 mmols NaOH. The C line then shows that the 5 mmols HAc reacts with 1.25 mmols NaOH to leave for the E line 3.75 for HAc, no NaOH left over, and 1.25 mmols of the salt., So at the end of the reaction, which is where we start plugging numbers into the equation, we have 3.75 mmols HAc and 1.25 mmols Ac^- (from NaAc). All of that isn't chemistry. it's just arithmetic. Let me point out something else while I'm here. Notice that the equation is
pH = pKa + log (NaAc/HAc). I gave you the way to convert mmols NaAc and mmols HAc to concentrations by (NaAc) = mmols/mL and you did all of that correctly. If you put all of what you did into the equation it would look like this. pH = 4.74 + log (1.25/62.5/3.75/62.5) = ? NOTICE that the volume (which is 62.5 in this case) cancels because it's the same number in the denominator of the numerator as well as the denominator of the denominator. In this equation the volumes WILL ALWAYS BE THE SAME SO THEY ALWAYS CANCEL which means you could have just used millimoles. Technically, that isn't right because concentration (i.e., mmols/mL) is what goes there BUT mathematically you don't need it. If your prof is picky (I was and the student got points subtracted when they did not use concentration). The students caught on fast to write the equation as
pH = pKa + log (1.25/V/3.75/V) = ??. Of course dividing by that V, which always cancels, make it right and they got full credit. Hope this helps.