Ionic equilibrium

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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24
eureka123 ·

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

can u finsih it now ??


i hope i didnt go wrong..correct me of i am wrong..[1]

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