Asked by mohit
Q.1 Calculate the extent of hydrolysis of 0.005M K2CrO4.If K2=5*10^-7 for H2CrO4.(It is essentially strong for first ionisation).
Answers
Answered by
DrBob222
CrO4^-2 + HOH ==> HCrO4^- + OH^-
Kh = kb = (Kw/k2) =
(HCrO4^-)(OH^-)/(CrO4^-2)
You know (CrO4^-2), Kw, and k2. (HCrO4^-) = (OH^-) = x.
Solve for x to determine (HCrO4^-), then extent hydrolysis = (HCrO4^-)/(CrO4^-2).
Check my work. Check my thinking.
Kh = kb = (Kw/k2) =
(HCrO4^-)(OH^-)/(CrO4^-2)
You know (CrO4^-2), Kw, and k2. (HCrO4^-) = (OH^-) = x.
Solve for x to determine (HCrO4^-), then extent hydrolysis = (HCrO4^-)/(CrO4^-2).
Check my work. Check my thinking.
Answered by
Sikander
Amother copied answer
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