Change in pH with addition of water?

Question

Calculate the change in pH (to three decimal places) that occurs in each of the solutions listed below as a result of a ten-fold dilution with water.

0.0500 M HCl

and also

0.0500 M CH3COOH

Answer 1

0.0500 M HCl diluted to 0.00500 M HCl
pH goes from -log(0.05) to -log(0.005) pH 1.30 to 2.30

0.0500 M HAc to 0.00500 M HAc
Ka=1.75 x 10^-5
HAc is my shorthand for acetic acid.
Ka=1.75 x 10^-5 = [H+][H+]/{[0.0500-[H+]}
Since acetic acid is only weakly ionized, I will assume that [0.0500] - [H+] is approximately [0.0500]. This avoids the need to solve a quadratic equation. (You can verify that the final answer won't change by solving the quadratic equation and comparing the answers.)
8.75x10^-7 = [H+]^2 and [H+] = 0.000935
pH=3.03
After dilution 10 fold-
1.75x10^-5 = [H+][H+]/[0.005]
8.75x10^-8 = [H+]^2 and [H+] =0.000295
pH=3.53

Answer 2

A strong acid goes up by 1.0 pH unit for a 10-fold dilution.
A weak acid goes up by 0.5.

Now work out the actual pH values of these acids, and you have your answer.

Answer 3

first the pH of 0.05M HCl is
- log (0.05) = 1.301
diluted by 10... or 0.005M HCl=
-log (0.005) = 2.301
so the change is 1

The pKa of acetic acid is 4.75696
the pH = pKa + log ([CH3COO+]^2/[CH3COOH])
solving pH=3.033
and for 10x dilution
pH = 3.542
so the difference is 0.509

Answer 4

pKa of CH3COOH needed.