[vmaggio]

What is the pH of a buffer made by mixing 50.0mL of 0.100M potassium hydrogen phthalate (KHP) with 13.6mL of 0.100M NaOH and diluting the mixture to 100.0mL with water? Ka2 = 3.6 x 10-6 for KHP.

A.3.25
B. 5.08
C. 5.51
D. 5.94




I am lost.




i started by getting mols of each.

mol/.050L = .005M
mol/.0136L= .00136M

-log(ka)= 5.44

i know i have to use the HH question, but i cant seem to figure out to correct method for the concentrations =\

[AWK]

You can calculate concentration of both reagents in the final volume, then do stoichiometry on concentrations


HP^- + OH^- =P^2- + H2O
0.05 M --------- 0.0136 --------- 0.0136
0.0364 (rest)
and use HH equation properly with these numbers (HP^- is an acid, P^2- - its conjugated base)