Determine the solubility of (a) `AgCI`, (b) `Fe(OH)_(3)`, (c) `Hg_(2)Br_(2)`, and (d) `Ag_(2)SO_(4)` from their solubility product constants give in table. Calculate the molarities of the individual ions and also the soubities of salts in `gL^(-1)`.

Explanation

a. For `AgCI, K_(sp) = 1.8 xx 10^(-10)`, hence
`AgCl_((S)) +aq hArr underset(S)(Ag^(oplus)(aq))+underset(S)(Cl^(Theta)(aq))`
`K_(sp) = [Ag^(o+)] [CI^(Θ)]`
`K_(sp) = S^(2)M^(2)`
i. `S//M = (1.8 xx 10^(-10)^(1//2) = 1//2) = 1.34 xx 10^(-5)`.
`[Ag^(o+)] = [CI^(Θ)], S = 1.34 xx 10^(-5) mol L^(-1)`
ii. `S` of `AgCI` in `gL^(-1) = 1.34 xx 10^(-5) xx 143.35 g mol^(-1)`
`= 1.92 xx 10^(-3) gL^(-1)`
b. For `Fe(OH)_(3): K_(sp) = 1.0 xx 10^(-38)`
`Fe(OH)_(2)(s)+hArr underset(S)(Fe^(3+))+underset(3S)(3 overset(Theta)(O)H)`
`:. K_(sp) = [Fe^(3+)] [overset(Θ)OH]^(3) = S.(3S)^(3) = 1^(1).3^(3)S^(4) = 27S^(4)M^(4)`
i. `S//M = ((1.0 xx 10^(-38))/(27))^((1/4)), S = 1.39 xx 10^(-10)`
ii. `[Fe^(3+)] = S = 1.39 xx 10^(-10) mol L^(-1)`
`[overset(Θ)OH] = 3S = 3 xx 1.39 xx 10^(-10)`
`= 4.17 xx 10^(-10) mol L^(-1)`
iii. `S` for `Fe(OH)_(3)` in `gL^(-1)`
`S = 1.39 xx 10^(-10) mol L^(-1) xx 106.85 g mol^(-1)`
`= 1.40 xx 10^(-18) gL^(-1)`
c. For mercurous bromide `(Hg_(2)Br_(2)): K_(sp) = 5.6 xx 10^(-23)`
Note: Mercurous ion exist as `Hg_(2)^(2+)`
`Hg_(2)Br_(2)(s)+aq hArr Hg_(2)^(2+) underset(S)(aq)+ underset(2S)(2Br^(Theta))(aq)`
`K_(sp) = [Hg_(2)^(2+)] [Br^(Θ)]^(2)`
`= (S) (2S)^(2) = 1^(1).2^(2).S^(3) = 4S^(3).M^(3)`
i. `S//M = ((5.6 xx 10^(-23))/(4))^((1/3)), S = 2.41 xx 10^(-8) mol L^(-1)`
ii. and `S` for `Hg_(2)br` in `gL^(-1)`
`S = 2.41 xx 10^(-8) mol L^(-1) xx 360.4 g mol^(-1)`
`= 8.68 xx 10^(-6) gL^(-1)`
iii. `[Hg_(2)^(2+)] = 2.41 xx 10^(-8)M`,
`[Br^(Θ)] = 2 xx 2.41 xx 10^(-8)M = 4.2 xx 10^(-8)M`
d. For silver sulphate `(Ag_(2)So_(4)): K_(sp) = 1.4 xx 10^(-5)`
`Ag_(2)SO_(4)(s)+aq hArr 2Ag^(oplus)underset(2S)((aq))+underset(S)(SO_(4)^(2-))(aq)`
`K_(sp) = [Ag^(o+)]^(2) [SO_(4)^(2-)]`
`= (2S)^(2) (S) = 2^(2). 1^(1).S^(3)M^(3) = 4S^(3) M^(3)`
i. `S//M = ((1.4 x 10^(-5))/(4))^((1//3)), S = 1.52 xx 10^(-2) mol L^(-1)`
ii. and `S` for `Ag_(2)SO_(4)` in `gL^(-1)`
`S = 1.52 xx 10^(-2) mol L^(-1) xx 311.8 g mol^(-1)`
`= 4.74 g L^(-1)`
iii. `[Ag^(o+)] = 2S = 2 xx 1.52 xx 10^(-2)`
`= 3.04 xx 10^(-2) mol L^(-1)`
`[SO_(4)^(2-)] = S = 1.52 xx 10^(-2) mol L^(-1)`