A soln of weak acid was titrated with NaOH. The equivalence pt. was reached when 36.12 ml of 0.10 (N) NaOH have been added. Now 18.06 ml of 0.1(N) HCl was added to the titrated soln.The pH was found to be 4.92. What's Ka of acid?
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1 Answers
eureka123
·2009-11-16 17:56:41
HA +NaOH ≈ ANa +H2O
36.12*0.1meq=3.612meq
ANa +HCl →HA +NaCl
3.612 1.806 0 0
0 1.806
so meq(HA)=1.806 and V=36.12mL+18.06 mL=54.18
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i hope i didnt go wrong..correct me of i am wrong..[1]