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(A)Find the solubility product of a saturated solution of Ag_(2)CrO_(4) in water at 298 K if the emf of the cell Ag|Ag+ (saturated. Ag_(2)CrO_(4) solution) || Ag+ (0.1 M) | Ag is 0.164 V at 298 K.(B) What will be the resultant pH when 200 mL of an aqueous solution of HCl (pH = 2.0) is mixed with 300 mL of an aqueous of NaOH (pH = 12.0) ? |
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Answer» Solution :(a) `E = 0.164 = -0.059 log[Ag^(+)]_(anode)/0.10` `[Ag^(+)]_(anode) = 1.66 xx 10^(-4) Ml` `[CrO_(4)^(2-)] = [Ag^(+)]/2 = 8.3 xx 10^(-5)M` `K_(SP) = [Ag^(+)]^(2)[CrO_(4)^(2-)] = (1.66 xx 10^(-4))^(2)(8.3 xx 10^(-5)) = 2.3 xx 10^(-12)` (b) pH of HCl = 2 `:. [H^(+)] = 10^(-2)M` Moles of `H^(+)` ions in 200 mL of `10^(-2) M` HCl solution` = 10^(-2)/1000 xx 200 = 2 xx 10^(-3`) Similarly, pH of NaOH = 12 `:. [H^(+)] = 10^(-12)M or [OH^(-)] = 10^(-2)M[:.[H^(+)][OH^(-)] = 10^(-14)M]` Moles of `[OH^(-)]` ion in 300 mL of `10^(-2) M` NaOH solution `= 10^(-2)/1000 xx 300 = 3 xx 10^(-3)` Total volume of solution after mixing = 500 mL Moles of `OH^(-)` ion left in 500 mL of solution `= (3 xx 10^(-3))-(2 xx 10^(-3)) = 10^(-3)` Solution `= 10^(-3)/500 xx 1000 = 2 xx 10^(-3) M` `pOH = - log(2 xx 10^(3)) = -log2 + 3log10 = -0.3 103 = 2.699` `:. pH = 14 - 2.699 = 11.301` |
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