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 (page 221). Determine also the molarities of individual ions.
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 (page 221). Determine also the molarities of individual ions.
(1) Silver chromate:
Ag2CrO4→2Ag++CrO2−4
Then,
Ksp=[Ag+]2[CrO2−4]
Let the solubility of Ag2CrO4 be s.
⇒[Ag+]2[CrO2−4]=s
Then,
Ksp=(2s)2.s=4s3⇒1.1×10−12=4s3.275×10−12=s3s=0.65×10−4M
Molarity ofAg+=2s=2×0.65×10−4=1.30×10−4MMolarity ofCrO2−4=s=0.65×10−4M
(2) Barium chromate:
BaCrO4→Ba2++CrO2−4
Then, Ksp=[Ba2+][CrO2−4]
Let the solubility of BaCRO4. be s.
So, [Ba2+]=s and [CrO2−4]=s ⇒Ksp=s2⇒1.2×10−10=s2⇒s=1.09×10−5M
Molarity of Ba2+ =Molarity if CrO2−4=s=1.09×10−5M
(3) Ferric hydroxide:
Fe(OH)3→Fe2++3PH−
Ksp=[Fe2+][OH−]3
Let s be the solubility of Fe(OH)3.
Thus, [Fe3+]=s and [OH−]=3s ⇒Ksp=s.(3s)3=s.27s3Ksp=27s41.0×10−38=27s4.037×10−38=s4.00037×10−36=s4⇒1.39×10−10M=S
Molarity of Fe3+=s=1.39×10−10M
Molarity of OH−=3s=4.17×10−10M
(4) Lead chloride:
PbCL2→Pb2++2Cl−KSP=[Pb2+][Cl−]2
Let KSP be the solubility of PbCl2
[Pb2+]=s and [Cl−]=2s
Thus, Ksp=2.(2s)2
=4s3⇒1.6×10−5=4s3⇒0.4×10−5=s34×10−6=s3⇒1.58×10−2M=S.1
Molarity of Pb2=s=1.58×10−2M
Molarity of chloride =2s=3.16×10−2M
(5) Mercurous iodide:
Hg2l2→Hg2++2l−Ksp=[Hg2+2]2[I−]2
Let s be the solubility of Hg2I2.
⇒[Hg2+2]=s and [I−]=2s
Thus, Ksp=s(2s)2⇒Kxp=4s34.5×10−29=4s31.125×10−29=s3⇒s=2.24×10−10M
Molarity of Hg2+2=s=2.24×10−10M
Molarity of I−=2s=4.48×10−10M
(1) Silver chromate:
Ag2CrO4→2Ag++CrO2−4
Then,
Ksp=[Ag+]2[CrO2−4]
Let the solubility of Ag2CrO4 be s.
⇒[Ag+]2[CrO2−4]=s
Then,
Ksp=(2s)2.s=4s3⇒1.1×10−12=4s3.275×10−12=s3s=0.65×10−4M
Molarity ofAg+=2s=2×0.65×10−4=1.30×10−4MMolarity ofCrO2−4=s=0.65×10−4M
(2) Barium chromate:
BaCrO4→Ba2++CrO2−4
Then, Ksp=[Ba2+][CrO2−4]
Let the solubility of BaCRO4. be s.
So, [Ba2+]=s and [CrO2−4]=s ⇒Ksp=s2⇒1.2×10−10=s2⇒s=1.09×10−5M
Molarity of Ba2+ =Molarity if CrO2−4=s=1.09×10−5M
(3) Ferric hydroxide:
Fe(OH)3→Fe2++3PH−
Ksp=[Fe2+][OH−]3
Let s be the solubility of Fe(OH)3.
Thus, [Fe3+]=s and [OH−]=3s ⇒Ksp=s.(3s)3=s.27s3Ksp=27s41.0×10−38=27s4.037×10−38=s4.00037×10−36=s4⇒1.39×10−10M=S
Molarity of Fe3+=s=1.39×10−10M
Molarity of OH−=3s=4.17×10−10M
(4) Lead chloride:
PbCL2→Pb2++2Cl−KSP=[Pb2+][Cl−]2
Let KSP be the solubility of PbCl2
[Pb2+]=s and [Cl−]=2s
Thus, Ksp=2.(2s)2
=4s3⇒1.6×10−5=4s3⇒0.4×10−5=s34×10−6=s3⇒1.58×10−2M=S.1
Molarity of Pb2=s=1.58×10−2M
Molarity of chloride =2s=3.16×10−2M
(5) Mercurous iodide:
Hg2l2→Hg2++2l−Ksp=[Hg2+2]2[I−]2
Let s be the solubility of Hg2I2.
⇒[Hg2+2]=s and [I−]=2s
Thus, Ksp=s(2s)2⇒Kxp=4s34.5×10−29=4s31.125×10−29=s3⇒s=2.24×10−10M
Molarity of Hg2+2=s=2.24×10−10M
Molarity of I−=2s=4.48×10−10M
