Question

26.20 mL of 0.102M NaOh is needed to neutralize 25.00mL of an unknown acid, HA. The pH at the equivalence point was measured to be 8.77. What is the Ka of this acid?

Answer 1

First we find out the no. of moles ofacid neutralised by NaOH as

moles of NaOH consumed = ( 26.20mL/ 1000L) * 0.102M solution.

= 2.67X10^-3 moles of NaOH

which is equal to the moles of HA at equivalence point = 2.67X10^-3 moles of HA

Let's assume that HA is a monoprotic acid, so the equation of interest is:

H+ + OH- -----> H2O

hence the initial concentration of HA = 2.67X10^-3 moles HA/ .025L =0.107 M

HA --> H+ + A-
at t=0 0.107M 0 0
at equivalence    0.107-x x x

now, we know that

10^-pH=[H+]

10^-8.77=1.70X10^-9M

K_a = x*x/ (0.107 - x)

and we know that [H+] = x = 1.70X10^-9M

Ka = 1.07 X10^-17