Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at `298K` from their solubility product constants given in Table `7.9`. Determine also the molarities of individual ions.

Answer

Silver chromate `S = 0.65 xx 10^(-4) M`; Molarity of `Ag^(+) = 1.30 xx 10^(-4) M`
Molarity of `CrO_(4^(2-)) = 0.65 xx 10^(-4) M`; Barium Chromate `S = 1.1 xx 10^(-5) M`; Molarity of
`Ba^(2+)` and `CrO_(4^(2-))` each is `1.1 xx 10^(-5) M`; Ferric Hydroxide `S = 1.39 xx 10^(-10) M`;
Molarity of `Fe^(3+) = 1.39 xx 10^(-10) M`; Molarity of `[OH^(-)] = 4.17 xx 10^(-10)M`
Lead Chloride `S = 1.59 xx 10^(-2) M`; Molarity of `Pb^(2+) = 1.59 xx 10^(-2) M`
Molarity of `Cl^(-) = 3.18 xx 10^(-2) M`; Mercurous Iodide `S = 2.24 xx 10^(-10) M`;
Molarity of `Hg_(2^(2+)) = 2.24 xx 10^(-10)M` and molarity of `I^(-) = 4.48 xx 10^(-10) M`

Explanation

(1) Silver chromate
`Ag_(2)CrO_(4)to2Ag^(+)+CrO_(4)^(2-)`
Then,
`K_(sp)=[Ag^(+)]^(2)[CrO_(4)^(2-)]`
Let the solubility of `Ag_(2)CrO_(4)` be s.
`rArr[Ag^(+)]2sand[CrO_(4)^(2-)]=s`
Then,
`K_(sp)=(2s)^(2).s=4s^(3)`
`rArr1.1xx10^(-12)=4s^(3)`
`.275xx10^(-12)=s^(3)`
`s=0.65xx10^(-4)` M
Molarity of `Ag^(+)=2s=2xx0.65xx10^(-4)=1.30xx10^(-4)` M
Molarity of `CrCO_(4)^(2-)=s=0.65xx10^(-4)` M
(2) Barium chromate:
`BaCrO_(4)toBa^(2+)+CrO_(4)^(2-)`
Then, `K_(sp)=[Ba^(2+)][CrO_(4)^(2-)]`
Let s be the solubility of `BaCrO_(4)`.
Thus, `[Ba^(2+)]=sand[CrO_(4)^(2-)]=s`
`rArrK_(SP)=S^(2)`
`rArr1.2xx10^(-10)=s^(2)`
`rArr s=1.09xx10^(-5)` M
Molarity of `Ba^(2+)` = Molarity of `CrO_(4)^(2-)=s=1.09xx10^(-5)` M
(3) Ferric hydroxide:
`Fe(OH)_(3)toFe^(2+)+3OH^(-)`
`K_(sp)=[Fe^(2+)][OH^(-)]^(3)`
Let s be the solubility of `Fe(OH)_(3)`.
Thus, `[Fe^(3+)]=s and [OH^(-)]=3s`
`rArrK_(sp)=s.(3s)^(3)=s.27s^(3)`
`K_(sp)=27s^(4)`
`1.0xx10^(-38)=27s^(4)`
`.037xx10^(-38)=s^(4)`
`.00037xx10^(-36)=s^(4)" "rArr1.39xx10^(-10)` M=S
Molarity of `Fe^(3+)=s=1.39xx10^(-10)` M
Molarity of `OH^(-)=3s=4.17xx10^(-10)` M
(4) Lead chloride:
`PbCl_(2)toPb^(2+)+2Cl^(-)`
`K_(SP)=[Pb^(2+)][Cl^(-)]^(2)` Let `K_(SP)` be the solubility of `PbCl_(2)`.
`[PB^(2+)]=sand [Cl^(-)]=2s`
Thus, `K_(sp)=s.(2s)^(2)`
`=4s^(3)`
`rArr1.6xx10^(-5)=4s^(3)`
`rArr0.4xx10^(-5)=s^(3)`
`4xx10^(-6)=s^(3)rArr1.58xx10^(-2)` M=S.1
Molarity of `PB^(2+)=s=1.58xx10^(-2)` M
Molarity of chloride `=2s=3.16xx10^(-2)` M
(5) Mercurous iodide:
`Hg_(2)I_(2)toHg^(2+)+2I^(-)`
`K_(sp)=[Hg_(2)^(2+)]^(2)[I^(-)]^(2)`
Let s be the solubility of `Hg_(2)I_(2)`.
`rArr[Hg_(2)^(2+)]=s and [I^(-)]=2s`
Thus, `K_(sp)=s(2s)^(2)rArrK_(sp)=4s^(3)`
`4.5xx10^(-29)=4s^(3)`
`1.125xx10^(-29)=s^(3)`
`rArrs=2.24xx10^(-10)` M
Molarity of `Hg_(2)^(2+)=s=2.24xx10^(-10)` M
Molarity of `I^(-)=2s=4.48xx10^(-10)` M