Posted by Joanna on Thursday, January 31, 2013 at 11:05am.
200 mL x 0.1M H2PO4^- = 20 millimols.
?mL x 0.5M NaOH =?
........H2PO4^- + OH^- ==> HPO4^- + H2O
I.......20........0..........0........0
add...............x...................
C.......-x.......-x...........x
E.......20-x......0...........x
pH = pK2 + log(base)/(acid)
Substitute into HH equation the E line from the ICE chart above and solve for x = millimoles OH needed.
Then M = millimols/mL. You know millimoles and M, solve for mL.
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