Question

Using a 0.20 M phosphate buffer with a pH of 6.5, you add 0.72 mL of 0.46 M NaOH to 58 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

I don't even know where to begin!! I just need a set up of the problem - without answers

Answer 1

You have two/three problems in one here. First I would determine the base and acid concentrations in the initial buffer. To do that use
pH = pKa + log (base)/(acid)
and solve for base/acid ratio.
Then you know acid + base = 0.2

Solve those two equation simultaneously to determine (H2PO4^-), the acid, and (HPO4^=), the base. I would then convert that to mols (or millimols) for the 58 mL sample. Let a = millimoles acid and b millimols base

............H2PO4^- + OH^- ==> HPO4^=
I.............a........0........b
add................mL x M = c........
C............ a-c.....-c........+c
E.............a-c......0........b+c

Then plug the E line back into a new HH equation and solve for the new pH.