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Calculate the pH of 0.10 M solution of NH_(4)Cl. The dissociation constant (K_(b)) of NH_(3) is 1.6xx10^(-5). |
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Answer» Solution :As `NH_(4) Cl` is a salt of weak BASE and strong acid, `K_(h)=(K_(w))/(K_(b))=(10^(-14))/(1.6xx10^(-5))=6.25 XX 10^(-10)` The hydrolysis REACTION will be `NH_(4)^(+) +H_(2)O hArr NH_(3)+H_(3)O^(+)` `{:(or,NH_(4)^(+),hArr,NH_(3),+,H^(+),,),("Initial conc.",cM,,,,,,),("At. eqm.",c-m,,x,,x,,):}` `K_(h)=([NH_(3)][H^(+)])/([NH_(4)^(+)])=(x xx x)/(c-x)~~ (x^(2))/(c) or x = sqrt(K_(h)xxc)=sqrt((6.25xx10^(-10))xx0.1)=7.9xx10^(-6)M` i.e., `[H^(+)] = 7.9 xx 10^(-6) M ` `:. pH = - log [H^(+)] = -log (7.9xx10^(-6))=6-0.90 = 5.10` Alternatively, applying the formula directly, `pH = 7 -(1)/(2) [ pK_(b) + log c ] = 7 - (1)/(2) [ - log K_(b)+log c ]` `=7-(1)/(2) [ - log (1.6xx10^(-5))+log 0.1 ]` `= 7 - (1)/(2) [ 5-0.02041) - 1] = 5.10` |
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