for the given volume and concentration of base & initial volume and pH of acid, find the Ka of the acid
Acid = 3-methylbutanoic acid
The conditions for your titration are:
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
To save typing let's call methylbutanoic acid = HA. Then
.....................HA + NaOH ==> NaA + H2O
millimoles NaOH in the titration = mL x M = 10.10 mL x 0.10M = 1.010
Therefore, millimoles HA = 1.010
Molarity of the HA = millimoles/mL = 1.010/10 = 0.1010. Then
.......................HA ===> H^+ + A^-
I.................0.1010...........0........0
C....................-x...............x........x
E................0.1010-x.........x........x
Ka = (H^+)(A^-)/(HA) = ?
Plug the E line into the Ka expression and solve for Ka. You will need to evaluate (HA), (H^+) and (A^-). You can do that from the pH of the acid = 2.76
pH = -log(H^+)
-2.76 = log(H^+). You can do this on your calculator. The answer is approximately 2E-3 and that is x. Evaluate (0.1010-x), (x) and plug those into the Ka expression and solve for Ka. Post your work if you get stuck.