# Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility p

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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.

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Correct 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
(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
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