# Homework Help: Titration of a weak base w/ a strong acid

1. Dec 4, 2012

### Chris L

1. The problem statement, all variables and given/known data

Determine the pH at the equivalence point after 20.00mL of a 3.75 M NaF(aq) solution is titrated with 3.25 M HI(aq)

2. Relevant equations

$k_a = \frac{[A^-][H^+]}{[HA]}$

3. The attempt at a solution

The first thing I noted was that

NaF $\rightarrow$ Na+(aq) + F-(aq)

since most sodium compounds dissolve almost completely, and

HI $\rightarrow$ H+(aq) + I-(aq)

since hydroiodic acid is strong. From there, the equilibrium expression

HF $\rightleftharpoons$ H+(aq) + F-(aq)

came to mind. Since the volume of HI(aq) added is one molar equivalent, and there are

$0.02 L (3.75 \frac{mol}{L}) = 0.075 mol$ of NaF(aq) in solution, I figured the volume of HI(aq) added must be $\frac{0.075 mol}{3.25 \frac{mol}{L}} = 0.0231 L$. The total volume is then $0.02 L + 0.0231 L = 0.0431 L$.

From the total volume, I found the new H+(aq) and F-(aq) concentrations to be $\frac{0.075 mol}{0.0431 L} = 1.74 M$; from my understanding, the initial concentrations of both of these entities should be the same because exactly 1 molar equivalence is being added and they are in a 1:1 ratio with each other.

Next, I created an ICE table with this information, along with the provided $k_a$ = 6.6 x 10-4 (I'm new to LaTeX so I put it inside a matrix b/c there's documentation for them in the FAQ, apologies):

$\begin{pmatrix} ? & HF & H^+ & F^-\\ I & 0 & 1.74 & 1.74\\ C & +x & -x & -x\\ E & x & 1.74 - x & 1.74 - x \end{pmatrix}$

Plugging this information into the expression for $k_a$,

$6.6*10^{-4} = \frac{(1.74 - x)^2}{x}$

which yields $x = 1.7064$ and so $[H^+] = 1.74 - 1.7064 = 0.0336 M$.

Finally, $pH = -log_{10}(0.0336) = 3.39$, which seems like a somewhat reasonable answer given the strength of the acid being added, but I still feel like there are errors in my methodology. I would be very thankful if anyone could point out my mistakes, or just tell me if everything looks okay!

Last edited: Dec 4, 2012
2. Dec 4, 2012

### Staff: Mentor

Note that it is obvious something is wrong - concentrations of both NaF and HI are very similar, there is no reason for tenfold difference in volumes, they have to be very similar.

3. Dec 4, 2012

### Chris L

Fixed the original post, I really have to get new batteries for my TI-89...I make too many mistakes with the windows calculator. Thanks for the response!

After fixing the numbers, the pH is now 3.39, although the highest pH given in the answers was 2.5 (it was multiple choice). The professor could just be trolling because it said "pick the closest value" but I think it's more likely that I made a second mistake.

4. Dec 4, 2012

### Staff: Mentor

Your previous answer was much closer to the correct one.

Notre that technically at equivalence point you have just HF solution of the known concentration...

5. Dec 4, 2012

### Chris L

I'm more concerned with the correct process than the right answer, my new answer being farther from the correct one doesn't bother me because I've fixed an error.

I'm not quite sure what you mean by the second statement; in the solution there exists Na+, H+, F-, and I-. From what I understand, the sodium and iodine ions shouldn't affect the pH, and the H+ and F- ions will form a certain concentration of HF, according to its $k_a$. Is it incorrect to say that the equivalence point is when the moles of HI added is equal to the moles of NaF in the original solution?

6. Dec 4, 2012

### Staff: Mentor

Not exactly. You add strong acid to protonate a weak base, and the reaction that takes place during titration is

H+ + F- -> HF

You don't add the acid to have free ions in the solution, you add it for the reaction to take place.

Correct. (They do change pH, but that's another story).

Compare the numbers you got - out of 1.74 M of F- present in the solution around 1.7M is protonated (even if the final answer is incorrect, orders of magnitude are OK). That means you have mostly HF, some of that dissociated, not partially protonated F-.

Now that I looked at your answer - minus log of 0.03 must be below 2 (minus log of 0.01 is 2, and we are talking about the number that is slightly higher than 0.01, so the minus log must be lower than 2). Check the final step.

7. Dec 4, 2012

### Chris L

Oh wow, I made another calculator mistake, -log10(0.0336) isn't 3.39, it's 1.474. Should that be the correct answer then?

8. Dec 4, 2012

### Staff: Mentor

Looks OK now.

9. Dec 4, 2012

### Chris L

Thank you for all of your help! I'm new to PF, is there a way I can give you a thumbs up or points or something?

10. Dec 4, 2012

### Staff: Mentor

No need for that, your "thank you" is enough