HC2H3O2 = HAc.
[C2H3O2]^- = Ac^-
It is easier to work in millimoles which I will do. Technically it isn't proper to use mols or millimols in the HH equation; however, since M = millimoles/mL and you have mmols/mL in numerator and denominator of the HH equation, the answer comes out the same because the mL unit cancels.
mmols HAc = mL x M = 25 x 0.01 = 0.25
mmols Ac^- = 25 x 0.11 = 2.75
mmols HCl added = 12 x 0.01 = 0.12
.........Ac^- + H^+ ==> HAc
I.......2.75....0.......0.25
add............0.12.............
C......-0.12..-0.12....+0.12
E......2.63......0......0.27
Plug the E line into the HH equation and solve for pH.
Check these figures, especially the decimal point. The procedure is right but I may have flubbed on the numbers. That's why you need to check them ALL.
12 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.11 M C2H3O2-. What is the pH of the resulting solution?
I understand to use the ice method, however, i do understand what to plug into the hh equation.
1 answer
1 answer