(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) ?

Explanation

(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`