1. Working in millimols. mmols NaoH added = 55 x 1.1 = 60.5
......HAc + OH^- ==> Ac^- + H2O
I.....400...0.........400
add........60.5...............
C..-60.5..-60.5.......+60.5
E...339.5...0........460.5
Convert the E line (in mmols) to M with M = mmols/total mL. Total mL = 100 + 55 = 155. Plug that into the Henderson-Hasselbalch equation and solve for pH. You are correct at 4.87
2.
Done the same way but you start with 400 mmols HAc and titrate exactly half of it. So you will form 200 mmols Acetate and you will have 200 mmols HAc remaining. Starting with those numbers you want to add 60.5 mmols NaOH and go from there. I get 5.01 for pH.
3. This one is a lot simpler because there is no buffer.
1.1 x 55 mL = 60.5 mmols NaOH added. (OH^-) = 60.5 mmols/155 mL = approx 0.4 but that's a close estimate. Convert that to pOH and then to pH. I get about 13.6
Two things here. You should see that with a buffered solution (both 1 and 2) there is little change in adding relatively high amounts of NaOH BUT with pure water with a pH = 7, the pH changes drastically (from 7 to 13.6) when the same amount of NaOH is added.
The second thing is, how did I know that the 400 mmols HAc would be end up at 200. First you know that it must be half and half between HAc and Ac^- but if you don't reason that out, you can do it this way. This is what you have in the titration.
.......HAc + OH^- ==> Ac^- + H2O
I......400...0.........0......0
add..........x.................
C......-x...-x.........+x
E.....400-x..0..........x
Now the problem tells you that you want the ratio to be 1:1 (which means you want 400-x = x)
Solve for x = 200 mmols NaOH must be added. That will be 200/1.1 = 181.82 mL of that 1.1 M NaOH solution so the final volume of THAT solution will be 100 mL + 181.81 mL + 55 mL = ?. I love these problems.
1)A solution prepared by adding 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate to 100.0 mL of water. The pH of this buffer solution is initially 4.74.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution
2) A solution prepared by adding 0.400 mol of acetic acid (pKa = 4.74) to 100.0 mL of water and then titrating with strong base until a buffer with a 1:1 buffer ratio is produced. The pH of this buffer solution is initially 4.74.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution
3) 100.0 mL of pure water.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution
For 1) I got 4.87 but I am having difficulties with the other two.
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