You have 800 mL of a 0.050M phosphate buffer, pH 6.5. You need to increase the pH of this buffer to 7.5 by using 6.00M NaOH. Determine the volume of NaOH needed. What will be the final concentration of the buffer?

Answer 1

I do these this way.
I used 6.34E-8 for k2.
6.5 = 7.2 + log (base)/(acid)
That gives me (base)/(acid) = 0.2 equation 1.
Equation 2 is acid + base = 0.05M
Solve these two equations simultaneously to obtain (acid) = 0.0417M
(base) = 0.00833M.

Then I change these to millimols.
800 x 0.0417 = 33.36 mmoles acid.
800 x 0.0833 = 6.667 mmols base
You want to add NaOH to make it 7.5
.........acid + OH^- ==> base + H2O
I........33.36....0.......6.667
.add.............x..............
C.........-x.....-x.........x
E.......33.36-x...x.........x

Substitute into the HH equation to obtain
7.5 = 7.2 + log base/acid
7.5 = 7.2 + log (6.667+x)/(33.36-x)
Solve for x = mmoles OH^- needed.
Since M = mmols/mL, plug in to find mL OH^- needed.

Post your work if you get stuck.