given volume and concentration of base & initial volume and pH of acid, find the Ka of the acid
Initial pH of acid: 2.76
Initial volume of acid: 10.0 mL
Concentration of NaOH: 0.10 M
Volume of NaOH required to reach equivalence: 10.10 mL
You omitted the stoichiometry for the acid and base so I will assume it is 1:1; i.e.,
HA + NaOH ==> NaA + H2O
millimoles NaOH = mL x M = 10.10 mL x 0.1 M = 1.010
millimoles HA must be 1.010 also in a 1:1 reaction.
M HA = millimoles/mL = 1.010/10.00 = 0.1010
pH = -log(H^+)
2.76 = -log(H^+)
(H^+) = 1.74E-3. Substitute for x in the below equation.
...................HA ---> H^+ + A^-
I..............0.1010.......0.........0
C..................-x...........x.........x
E.............0.1010-x......x.........x
Ka = (H^+)(A^-)/(HA) = (x)(x)/(0.1010-x)
You know the value of x from above. Substitute and solve for Ka. Should be around 10^-5 or so. Post your work if you get stuck.