HC2H3O2 = HAc and C2H3O2^- = Ac^-
First the addition of NaOH.
..................HAc + OH^- ==> Ac^- + HOH
I.................0.1.......0.................0.1.................
add....................0.015..................................
C............-0.015 ...-0.015.............+0.015................
E..............0.085.......0...................0.115
(HAc) = 0.085/volume what ever that is. It's the same for (Ac^-) and that is (0.115/volume). Note the volumes cancel, whatever that is, so the equation looks like this. pH = pKa + log (base/acid) which is the Henderson-Hasselbach equation
pH = 4.74 + log (0.115/0.085) and solve for pH.
For part b, the addition of HCl it is done the same way as the addition of NaOH except the equation is a little different since you're adding acid and not base AND the volumes are given so you can calculate the concentration to put into the HH equation. Just remember though that the volume cancels so you really don't need to do the actual concentration calculation; i.e., you can use moles. I'll leave that calculation for you. Post your work if you get stuck. Here is the new reaction you need.
..........................Ac^- + H^+ ==> HAc
A 1.0L buffer solution contains 0.100 mol
of HC2H3O2and 0.100 molof NaC2H3O2
. The value of Ka
for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution upon the addition of 0.015 mol NaOH to the original buffer
Calculate the pH of the solution upon the addition of 10.0 mL of 1.00 M HCl to the original buffer.
1 answer
1 answer