(a) Addition of 0.00 mL of strong base:

The solution contains only 0.100 M strong acid.  Strong acids completely ionize in aqueous solution.

HBr_(aq)  +  H₂O_(l)  ----->  H₃O⁺_(aq)   +  Br^-_(aq)

Therefore, [H₃O⁺] = 0.100 M which gives a pH = 1.000, ( recall:  pH = - log[0.100]).

(b)  Addition of 10.00 mL of strong base:  (Excess strong acid present)

First solve for the final concentrations at the end of the strong acid/strong base reaction:

HBr_(aq)  +  KOH_(aq)  ----->  H₂O_(l)   +  KBr_(aq)

The net ionic equation for this reaction can be written and a reaction table set up (omitting spectator ions):

H₃O⁺_(aq)  +  OH^-_(aq)  ----->  2 H₂O_(l)         K = (1 / K_w) = 10¹⁴, goes to completion

moles initial   0.00500          0.00200                 ~     

moles final    0.00300               0                        ~          Total Volume = (50.00 + 10.00) mL = 0.06000 L

Therefore, [H₃O⁺]_final = [0.00300 moles / 0.06000 L] = 0.0500 M which gives a pH = 1.301.

(c)   Addition of 24.90 mL of strong base:

First solve for the final concentrations at the end of the strong acid/strong base reaction:

H₃O⁺_(aq)  +  OH^-_(aq)  ----->  2 H₂O_(l)

moles initial   0.00500           0.00498             ~     

moles final    0.00002              0                       ~              Total Volume =  0.07490 L

Therefore, [H₃O⁺]_final = [2 x 10^-5 moles / 0.07490 L] = 3 x 10^-4 M which gives a pH = 3.5.

(d)   Addition of 25.00 mL of strong base:  (The Equivalence Point)

First solve for the final concentrations at the end of the strong acid/strong base reaction:

H₃O⁺_(aq)  +  OH^-_(aq)  ----->  2 H₂O_(l)

moles initial   0.00500       0.00500                 ~     

moles final          0                   0                       ~              Total Volume =  0.07500 L

Therefore, [H₃O⁺]_final = 1.0 x 10^-7 M which is only from the autoionization of water,  pH = 7.00.

(e)   Addition of 25.10 mL of strong base:  (Excess strong base added)

First solve for the final concentrations at the end of the strong acid/strong base reaction:

H₃O⁺_(aq)  +  OH^-_(aq)  ----->  2 H₂O_(l)

moles initial   0.00500           0.00502             ~     

moles final         0                  0.00002              ~              Total Volume =  0.07510 L

Therefore, [OH^-]_final = [2 x 10^-5 moles / 0.07510 L] = 3 x 10^-4 M

and the [H₃O⁺]_final = (K_w / [OH^-]) = (1.0 x 10^-14  /  3 x 10^-4 ) = 3 x 10^-11 M which gives a pH = 10.5.

(f)   Addition of 30.00 mL of strong base: 

First solve for the final concentrations at the end of the strong acid/strong base reaction:

H₃O⁺_(aq)  +  OH^-_(aq)  ----->  2 H₂O_(l)

moles initial   0.00500           0.00600             ~     

moles final         0                  0.00100              ~              Total Volume =  0.08000 L

Therefore, [OH^-]_final = [1.00 x 10^-3 moles / 0.08000 L] = 1.25 x 10^-2 M

and the [H₃O⁺]_final = (K_w / [OH^-]) = (1.0 x 10^-14  /  1.25 x 10^-2 ) = 8.00 x 10^-13 M which gives a pH = 12.10.

(g)   Addition of 40.00 mL of strong base: 

First solve for the final concentrations at the end of the strong acid/strong base reaction:

H₃O⁺_(aq)  +  OH^-_(aq)  ----->  2 H₂O_(l)

moles initial   0.00500           0.00800             ~     

moles final         0                  0.00300              ~              Total Volume =  0.09000 L

Therefore, [OH^-]_final = [3.00 x 10^-3 moles / 0.09000 L] = 3.33 x 10^-2 M

and the [H₃O⁺]_final = (K_w / [OH^-]) = (1.0 x 10^-14  /  3.33 x 10^-2 ) = 3.00 x 10^-13 M which gives a pH = 12.52.