Question

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous hydrofluoric acid requires 30.00 mL of 0.400 M NaOH? Ka = 6.76 × 10-4 for HF.

Answer 1

Find the pH when equal amounts of NaOH and HF have reacted.

Answer 2

8.25?

Answer 3

Working now. I'll post when I get it figured out. 8.25 is one of the choices, so it seems likely.

Answer 4

Sigh... I wish I took notes last year ;/ I'm not that confident in my chem skills anymore...

Answer 5

How did you go about getting your answer?

I've got...\[NaOH+HF \rightleftharpoons NaF + H_2O \]There are going to be 0.012 moles of NaOH, so at the equivalence point there needs to be 0.012 moles of HF. At this new concentration of solution, the concentration of HF at the equivalence point needs to be 0.218M.

I'm not sure what I need to do from here. Is an ICE chart involved?

Answer 6

Yeah, you'll have to use the Ice chart.\[_p k _b = \frac {x^2}{0.218 - x}\]x is your [OH-] so after that, just find the pOH and then pH.

Answer 7

Why are we using the pKb, instead of the Ka?

Answer 8

The Ka here represents...

\[K_a=\frac{[H^+][F^-]}{[HF]}\]Which is the equation that the ICE chart you used set up used, correct?

Answer 9

\[F^- + H_2 O \rightarrow HF + OH^-\]

Answer 10

Why? We know the concentration of HF, right? So why not use...\[HF(aq)+H_2O(l) \leftrightharpoons F^-(aq)+H^+(aq)\]

Answer 11

No, you know the concentration of NaF not HF at the equivalence point. At the equivalence point, NaOH has neutralized all the HF.

Answer 12

You have to find the new equilibrium established with the 0.218 M NaF.

Answer 13

My wording is sort of off, but hopefully you get the drift ;)

Answer 14

So basically, at the equivalence point, all reactant has been consumed in this equation?

The equation...\[NaOH+HF \rightarrow NaF + H_2O\]

And then, since we know how much of the neutral salt, NaF, was produced, it's essentially just a pH calculation of the NaF neutral salt in solution?

Answer 15

Yes =)

Answer 16

Its like buffers.

Answer 17

Ah.

So we're just seeing how much fluoride is going to associate back into HF?

If it were a salt composed of a weak acids conjugate base and a weak bases conjugate acid, would the problem be more complicated?

Answer 18

Well, if you have a weak acid & a weak base, then its a buffer. If strong acid + weak base or strong base + weak acid, thats just normal titration.

Answer 19

In other words, this problem is only so simple because the sodium cation is a weak acid, so it won't associate with hydroxide very much?

Answer 20

I see. Wonderful! (:

Titrations have always been confusing to me :P. Now I think I understand them at least a little bit better after this. Thank you!

Answer 21

Na+ is a negligible acid, it doesn't have protons nor does it have an affinity to make coordinate covalent bonds :P

Answer 22

Buffers are still simpler in my opinion. Titration problems are not pretty when your dealing with di or triprotic acids.

Answer 23

I still don't understand, and I am not sure where I am making the disconnection

moles of OH = .030 L * .4 M = .0012 moles OH-

concentration of F- = .0012/(.030 + .025) = 2.18E-2 M

and thats where i go blank... Im not sure what you do after this.

Answer 24

Actually moles of OH = 0.03L * 0.4M = 0.012 moles of OH-

The concentration of F- = 0.012/(0.03 + 0.025) = 0.218

For the ICE equation use:
\[F ^{-} + H _{2}O \rightarrow HF + OH ^{-}\]
0.218 0 0
-x +x +x
0.218-x +x +x

Convert the Ka to Kb by using the formula: \[Ka \times Kb = Kw\]
\[6.76 \times 10^{-4} (Kb) =1.0 \times 10^{-14}\]
divide each side by \[6.76 \times 10^{-4}\]
\[Kb = 1.0 \times 10^{-14} \div 6.76 \times 10^{-4} = 1.48 \times 10^{-11}\]

Then use the equation:\[Kb = [HF] \times [OH ^{-}] \div [F ^{-}]\]
\[1.48 \times 10^{-11} = x ^{2} \div 0.218\]
You can leave "-x" out of the equation because it is so small
\[3.22 \times 10^{-12} = x ^{2}\]
take the square root on each side
\[\sqrt{3.22 \times 10^{-12}} = \sqrt{x ^{2}}\]
\[1.796 \times 10 ^{-6} = x\]
find the pOH by taking the negative log of x
\[pOH = -\log_{1.796 \times 10^{-6}} = 5.75\]
Find the pH
\[pH = 14 - 5.75 = 8.25\]

pH = 8.25