HBr + NaOH ==== Na+ + Br- + H2O
So the 1.2 x 10 -4 M OH- in NaOH will be completely neutralized by the H+ in HBr, which makes the concentration of H+ in the solution at this point:
[H+]=0.0893 -1.2 x 10 -4 =0.08918 M
Now we can think about the buffer.
HC2H3O2 ==== H+ + C2H3O2- pKa=4.76
So when the above reaction reaches its balance,
[H+][C2H3O2-]/[HC2H3O2 ]=10^-4.76=1.74 x 10 -5
[H+][C2H3O2-]/[HC2H3O2 ]=0.08918 x 0.05 /0.15 = 0.02973 >> 1.74 x 10 -5
So the reaction will move in the left direction. Assuming a M H+ is converted to HC2H3O2 to reach the final balance, we can now get the following equation
(0.08918-a) (0.05-a) / (0.15 + a) = 1.74 x 10 -5
(0.08918-a) (0.05-a) = 1.74 x 10 -5 (0.15 + a)
Since a has to be not larger than 0.05, so after the calculation you will find that a=0.05, which means all C2H3O2- is converted to HC2H3O2 at the end.
Therefore the final concentration of H+
[H+] = 0.08918-0.05 = 0.03918 M
pH = 1.4
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