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Explain about the hydrolysis of salt of strong acid and weak base . Derive K_b and pH for that solution. |
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Answer» SOLUTION :(i) Consider a REACTION between STRONG acid HCl and a weak base `NH_4OH` to produce a salt `NH_4Cl` and water `HCl(aq)+NH_4OH(aq) leftrightarrowNH_4Cl(aq)+H_2O_(l)` `NH_4Cl_((aq)) to NH_4^(+) (aq) +CL^(-) (aq)` (ii) `NH_4^(+)` is a strong conjugate acid of the weak base `NH_4OH` and its has a tendency to react with `OH^(-)` from water to produce unionised `NH_4` as below `NH_4^(+) (aq) toNH_4OH_((aq))+H^(+) (aq)` (iii) There is no such tendency shown by `Cl^-` and therefore `[H^+] gt [OH^-]` the solution is acidic and the pH is less than 7. (iv) In the salt hydrolysis of strong base and weak acid, we have to derive a relationship between `K_a and K_b` as (v) `K_h.K_b=K_w` `K_h=h^2C and [H^+]=sqrt(kh.C)` `therefore[H^+]=sqrt((K_w.C)/K_b)` `pH=-log[H^+]` `=-log[(K_w.C)/K_b]^(1//2)` `=-1/2logK_w-1/2log C+1/2 log K_b` `pH=7-1/2 pK_b -1/2 log C` |
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