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If the equilibrium constant for the reaction of weak acid HA with strong base is `10^(9)`, then pH of `0.1M` Na A is: |
Answer» Correct Answer - 9 `HA+NaOH rarr NaA+H_(2)O` or `HA+OH^(-)rarr A^(-)+H_(2)O` `K_(eq.)=10^(9)=([A^(-)][H_(2)O])/([HA][OH^(-)])` Also `HAhArrH^(+)+A^(-)` `K_(a)= ([H^(+)][A^(-)])/([HA])` `:. (K_(eq.))/(K_(a))=(1)/(K_(w))` or `K_(a)=10^(9)xx10^(-14)=10^(-5)` Thus, for `A^(-)H_(2)OhArrHA+OH` `[OH^(-)]= c.h=c. sqrt((K_(H))/(c ))= sqrt((K_(w).c)/(K_(a)))` `=sqrt((10^(-14)xx-0.1)/(10^(-5)))=10^(-5)` `:. [H^(+)]=10^(-9) implies pH =9` |
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