need help with Buffer calculation please.?

Question

30mL of 0.1M acetic acid with 15mL of 0.1M sodium hydroxide...

1) what's the pH of the buffer solution
2) 5mL of NaOH is added. what's the pH of the new buffer solution..

someone please help me.
thank you..

Answer 1

The approach given by the ghost whisperer above is generally correct, except that the Henderson-Hasselbalch formula is

pH = pKa + log[A-]/[HA]

If you use pKa (which is the -log of Ka) you do not take the log again. Also, note that it is [conjugate base] over [acid]; it didn't matter in the first part, because the concentrations were the same.

The result should be 4.74 + log(1) = 4.74, not the log of 4.74

For the second part, you are adding 5mL of 0.1M NaOH, or 0.005*0.1 = 0.0005 moles NaOH. This will react with 0.0005 moles of remaining HAc, reducing the amount by 0.0005 moles. In the first part, you calculated that you had 0.0015 moles of HAc remaining after the reaction, so now you have 0.0015 - 0.0005 = 0.0010 moles of HAc in a total 50ml of solution (30 + 15 + 5) so [HAc] = 0.0010/0.05 = 0.02M. We have increased the moles of NaAc by the same amount (0.0005 moles); that was 0.0015 and is now 0.0020 moles in 50mL, so [Ac] = 0.002/0.05 = .04M. Now apply the Henderson-Hasselbalch equation:

pH = pKa + log0.004/0.002 = 4.74 + log(2) = 4.74 + 0.301 = 5.04.

(P.S. I am assuming that ghost whisperer's value of pKa for the HAc reaction is correct. EDIT: I just checked that: wikipedia gives Ka for acetic acid = 4.76; close enough, I guess.)

Answer 2

1) The reaction is CH3COOH + NaOH -> CH3COO-Na+ + H2O

First, calculate what is the limiting agent.
amt of acetic acid=30/1000*0.1=0.003mol
amt of sodium hydroxide=15/1000*0.1=0.0015mol
So NaOH is the limiting reagent.
Therefore, amt of acetate formed is 0.0015mol

amt of acetic acid remaining(unreacted)=0.0015mol
conc. of acetic acid=amt/vol
=0.0015/(45/1000)=0.333mol/dm3
conc. of acetate=amt/vol
=0.0015/(45/1000)=0.333mol/dm3

pH of buffer solution
=lg (pKa)+lg ([acid]/[salt])
=lg(4.74)+lg (0.333/0.333)
=0.675778341
=0.676(3s.f.)

2)For this part, caiculate the amount of acid remaining and acetate formed. The remaining steps are similar to above.

Good luck. ;P