Destroyer General Chem Q201

[Meas]

HOw many ml of 0.120M KOH must be added to 60ml of 0.2M HF to produce a solution with a pH of 3.3? (pKa of HF is 3.3)

Ans: 5ml

- THe answer key shows that c1v1 = c2v2 eqn can be used.
- When I first approach this question, I quickly use the pH to find out how much [H+] is needed.

Please help! Thanks in advance!

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[nze82]

HOw many ml of 0.120M KOH must be added to 60ml of 0.2M HF to produce a solution with a pH of 3.3? (pKa of HF is 3.3)

Ans: 5ml

- THe answer key shows that c1v1 = c2v2 eqn can be used.
- When I first approach this question, I quickly use the pH to find out how much [H+] is needed.

Please help! Thanks in advance!

Notice how pKa = pH. This is the characteristic of the half point. Why?
Remember that:
pH = pKa + log [A-]/[HA]

At half point [A-] = [HA], so log [A-]/[HA] = 1, so:
pH = pKa + log1 = pKa

So, in this example, you just need to find the volume of KOH added to reach the half point. How?

Use M1V1 = M2V2

V1(0.12M) = (60ml)(0.2M)
V1 = 100ml

However, this is the volume of KOH added to reach the equivalence point; whereas, we're looking for the volume at the half-point. So, just divide this volume by 2.

Ans: 100ml/2 = 50ml

 

[WOAHHI]

anyone know if this question was a typo in destroyer? the answer says 5 mL not 50 mL

 

[whawha]

What nze82 did is perfectly correct, and I think the answer should be 50 mL, not 5 mL as given in the solution.

Here is how I approached this problem before I looked at the solution.

# of moles of [H+] in HF = 0.2 * 0.06 = 0.012

At pKa (given to be 3.3), it will lose 50% of the protons, which means it will have 0.006 moles of [H+]

That means, a certain amount of OH- needs to be used to get reacted with 0.006 moles of [H+] of the 0.012.

x mL of KOH = 0.006 moles / 0.12 M = 0.05 L = 50 mL

Hope this helps.

 

[Marmaduke]

yeah I think it was a typo. the error was at the (60ml)(.2) = 1.2 instead of the real anser of 12.