InterviewSolution
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InterviewSolution
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How much the concentration of `Ag^(o+)`ions in a saturted solution of `AgCI` diminish if such an amount of `HCI` is added to it that the concentration of `CI^(Θ)` ions in the solution becomes equal to `0.03M`? Also find the amount of `AgCI` precipitated at the given concentration. `K_(sp)` of `AgCI = 1.8 xx 10^(-10)`. |
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Answer» `HCI` is added to a solution containing `Ag^(o+)` ions in saturated solution. First find the concentration of `Ag^(o+)` ion in this solution `AgCl(s) hArr Ag^(oplus) + Cl^(Theta)` `rArr K_(sp) [Ag^(o+)] [CI^(Θ)] = x^(2)` where `x` is solubility of `AgCI` in mol `L^(-1)` `rArr [Ag^(o+)] = sqrt(K_(sp)) = sqrt(1.8 xx 10^(-10)) = 1.34 xx 10^(-5)M` When `HCI` is added, the ionic product of `AgCI` approaches the `K_(sp)` value of `AgCI`, the precipiation of `Ag^(o+)` ions will occur. As ionis product increase (i.e., become greater than `K_(sp)` value), and appreciable amount of `AgCO` precipitates out, and precipitation continues till ionic product equals solubility product `(K_(sp))`. Ionic product `= [Ag^(o+)] [CI^(Θ)] = K_(sp)` `rArr [Ag^(o+)] = (K_(sp))/([CI^(Θ)]) = (1.8 xx 10^(-10))/((0.03)) = 6.0 xx 10^(-9)M` Now this is the amount of `Ag^(o)` ions left un-precipitated `rArr [Ag^(o+)]` diminishes in the soolution by `(6.0 xx 10^(-9))/(1.34 xx 10^(-5)) = ((1)/(2233))` times The concentration of `AgCI` precipitated out of the solution `= [Ag^(o+)]_("initial") - [Ag^(o+)]_("left")` `= 1.34 xx 10^(-5) - 6.0 xx 10^(-9)M` It means almost whole of `AgCI` is precipitated out of the solution at `[CI^(Θ)] = 0.03M`. |
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