7.67 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.

(1) Silver chromate

Ag2CrO42Ag++CrO42-

Then,

Ksp=Ag+2CrO42-

Let the solubility of Ag2CrO4 be s.

Ag+2s and CrO42-=sThen,Ksp=2s2.s=4s31.1×10-12=4s3.275×10-12=s3s=0.65×10-12=s3s=0.65×10-4 M

Molarity of Ag+         =2s=2×0.65×10-4=1.30×10-4 M         

Molarity of CrO42-            =s=0.65×10-4 M

(2) Barium chromate:

BaCrO4Ba2++CrO42-Then, Ksp=Ba2+CrO42-

Let the solubility of BaCrO4. be s.

So,

Ba2+=s and CrO42-=s       Ksp=s2                                                1.2×10-10=s2                                                s=1.09×10-5 M

Molarity of Ba2+=Molarity of CrO42-=s=1.09×10-5 M

(3) Ferric hydroxide:Fe(O)3Fe2++3OH-Ksp=[Fe2+][OH-]3Let s be the solubility of  Fe(OH)3 Thus, [Fe3+]=s and [OH-]=3s  Ksp=s.(3s)3 =s.27s3Ksp=27s4.037×10-38=s4.00037×10-36=s4 1.39×10-10 M=SMolarity of Fe3+=s=1.39×10-10MMolarity of OH-=3s=4.17×10-10 M

(4) Lead chloride:PbCI2pb2++2CI-Ksp=[pb2+][CI-]=2sThus, Ksp=s.(2s)2=4s31.6×10-5=4s30.4×10-5=s34×10-6=s31.58×10-2M=S.IMolarity of PB2+=s=1.58×10-2MMolarity of chloride = 2s=3.16×10-2M(5) Mercurous iodide: Hg2I2Hg2++2I-

Ksp=Hg22+2I-2

Let be the solubility of Hg2l2.

Hg22+=s and l-=2sThus, Ksp=s2s2Ksp=4s34.5×10-29=4s31.125×10-29=s3s=2.24×10-10 M

Molarity of Hg22+=s=2.24×10-10 M

Molarity of I-=2s=4.48×10-10 M