'A solution is prepared by titrating a 100.0 mL sample of 0.10 M HF (Ka = 7.2 x 10^-4) with 0.1 M NaOH. What is the pH after 20.0 mL of the 0.10 M NaOH is added?'
I'm a little bit confused but here's my ICE chart (in moles):
HF________OH-_____F-
1.0_______0.002___0
-0.002____-0.002__+0.002
0.998______0______0.002
So if the total volume is now 120 mL (=0.12 L), that means the concentration of HF is about 8.32 M and the concentration of F- is about 0.017. I found pKa using -log(Ka) and then I plugged everything into the pH = pkA + log [A-]/[HA] equation, but that gave me 0.424, and I know that's not correct. Where did I go wrong?
Thanks!
You're on the right track but I don't like your ICE chart. That 1.0 for HF isn't right. Where did you go wrong. Two places. I think the HF is (0.1*0.1/0.12) = 0.0833. I agree with the 0.017(I used 0.01667) for F^-. The second error is that you didn't subtract the added NaOH from the HF beginning; i.e., 0.0833 - 0.0167 = ?. Then plug those new numbers into the Henderson-Hasselbalch equation. I get something like 2.57. Hope this helps. Here is the way I do it and I think it makes it much simpler; however, some profs frown on it. To be honest about it, I always took points off when my students did it but the answer comes out the same. Work in millimols.
mmols HF = 100 x 0.1 = 10
mmols OH added = 20 x 0.1 = 2
........HF + OH^- ==> F^- + H2O
I.......10....0.......0......0
add...........2...............
C......-2....-2.......+2
E.......8.....0........2
pH = pKa + log (b)/(a) = 3.17 + log 1/4 = about 2.57.
Here is the technical problem. The HH equations says that (base) and (acid) are in M = mols/L and my shortcut I use is in mols and not mol/L (actually mmols and not mmols/mL). So why does it come out the same? Because the volume in a titration is ALWAYS the same for base and acid; i.e., if we write it right we have pH = 3.17 + log (2/0.12)/(8/0.12) and the 0.12 cancels (every time) so I just remember this.
pH = pKa + log (mmols base/mL)/(mmols acid/mL) = ? The mL cancels every time and mmols stay and the answer always works. It also saves the time of dividing that 2/0.12 and 8/0.12.
To find the pH after adding 20.0 mL of 0.1 M NaOH to the 100.0 mL sample of 0.10 M HF, you need to consider the reaction between HF and NaOH. HF is a weak acid, and NaOH is a strong base, so they will react to form the corresponding salt and water.
The balanced chemical equation for the reaction is:
HF(aq) + NaOH(aq) -> NaF(aq) + H2O(l)
Initially, you have 100.0 mL of 0.10 M HF. After adding 20.0 mL of 0.10 M NaOH, the reaction occurs between HF and NaOH, resulting in the formation of NaF and water. The moles of NaOH used can be found by multiplying the volume (20.0 mL) by the molarity (0.10 M) of NaOH, which gives:
moles of NaOH used = 20.0 mL * 0.10 M = 0.002 moles
Since HF and NaOH react in a 1:1 ratio according to the balanced equation, 0.002 moles of HF will be neutralized. The remaining moles of HF can be calculated as follows:
moles of HF remaining = initial moles of HF - moles of NaOH used
= 0.1 M * 0.100 L - 0.002 moles
= 0.01 moles
Now, let's calculate the moles of F- formed. Since the reaction is 1:1, the moles of F- formed will also be 0.002 moles.
Next, we need to calculate the final volume of the solution after the addition of NaOH. Initially, you have 100.0 mL of HF, and after adding 20.0 mL of NaOH, the total volume becomes 120.0 mL or 0.12 L.
To find the concentration of HF and F-, divide the moles by the final volume:
[HF] = moles of HF / final volume
= 0.01 moles / 0.12 L
= 0.083 M (approximately)
[F-] = moles of F- / final volume
= 0.002 moles / 0.12 L
= 0.017 M (approximately)
Now, we can calculate the pH using the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of F- and [HA] is the concentration of HF.
The pKa value for HF can be found by taking the negative logarithm of Ka:
pKa = -log(Ka)
= -log(7.2 x 10^-4)
= 3.14 (approximately)
Now, plug in the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= 3.14 + log(0.017/0.083)
= 3.14 - 0.54
= 2.60 (approximately)
Therefore, the pH after adding 20.0 mL of 0.10 M NaOH is approximately 2.60.
So, where you went wrong in your calculations was in assuming that the concentration of HF would be 8.32 M and the concentration of F- would be 0.017 M after the reaction. You needed to consider the neutralization reaction between HF and NaOH, as well as the change in volume caused by adding NaOH.