millimols HAc = 75 mL x 0.1M 7.5.
mmols NaAc = 75 mL x 0.2M = 15.
9.5 mL x 0.1M HCl = 0.95 mmols.
........Ac^- + H^+ ==> HAc
I.......15.....0........7.5
add...........0.95............
C....-0.95..-0.95.......+0.95
E.....14.05...0........8.45
Substitute the E line into the HH equation and solve for H.
Calculate the pH of 75 ml of the buffer composed of .1M HC2H3O2 with .2M NaC2H3O2; to which 9.5 ml of .1M HCl is added.
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