Let's simplify this. HC7H5O2 is benzoic acid = HBz = 0.1 M
Ca(C7H5O2)2 is calcium benzoate = Ca(Bz)2 = 0.2 M
You have mols HBz = M x L = 0.1 x 5 = 0.5 mol = acid
and Bz^- from Ca(Bz)2 = 0.4 x 5 = 2 mols = base
After adding 10 mL of 5 M NaOH = 0.01 x 5 = 0.05 mols then
.................HBz + OH^- ==> Bz^- + H2O
I................0.5..........0...........2.....................
add........................0.05.........................................
C............-0.05......-0.05......+0,05
E.............0.45...........0.......2.05
Plug the E values into the Henderson-Hasselbalch equation and solve for pH.
Post your work if you get stuck.
A 5.00 L solution has [HC7H5O2] = 0.100 M and [Ca(C7H5O2)2] = 0.200 M . What is the pH of the solution after 10.0 mL OF 5.00 M NaOH is added? Assume the volume of the solution does not change .Ka for HC7H5O2 = 6.3x10^-5.
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