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?
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
Ka = x*x/ (0.107 - x)
and we know that [H+] = x = 1.70X10-9M
Ka = 1.07 X10-17
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