Problem 90P

Complete the table. Assume that all concentrations are equilibrium concentrations in moles per liter, M.

Compound |
[Cation] |
[Anion] |
Ksp |

CdS |
3.7 × 10−15 |
3.7 × 10−15 |
_____ |

BaF2 |
_____ |
7.2 × 10−3 |
1.9 × 10−7 |

Ag2SO4 |
2.8×10−2 |
_____ |
1.1 × 10−5 |

Solution: Here, we are going to complete the given table.

Step1:

Solubility product of any substance is the product of the concentrations of the dissolved ions raised to the power of their stoichiometric coefficients.

Consider the ionization of a substance AB,

AB A+ + B-

Then, the solubility product, Ksp = [A+][B-]

\Step2:

Compound |
[Cation] |
[Anion] |
Ksp |

CdS |
3.7 x 10-15 |
3.7 x 10-15 |
1.369 x 10-29 |

BaF2 |
3.665 x 10-3 |
7.2 x 10-3 |
1.9 x 10-7 |

Ag2SO4 |
2.8 x 10-2 |
1.403 x 10-2 |
1.1 x 10-5 |

- The ionization of CdS takes place in the following manner:

CdS Cd2+ + S2-

Solubility product for the reaction, Ksp = [Cd2+][S2-]

= (3.7 x 10-15)(3.7 x 10-15)

= 1.369 x 10-29

b) The ionization of BaF2 takes place in the following manner:

BaF2 Ba2+ + 2F-

Solubility product for the reaction, Ksp = [Ba2+][F-]2

Therefore, [Ba2+] = Ksp / [F-]2

= (1.9 x 10-7) / (7.2 x 10-3)2

= 0.03665 x 10-1

= 3.665 x 10-3

c) The ionization of Ag2SO4 takes place in the following manner:

Ag2SO4 2Ag+ + SO42-

Solubility product for the reaction, Ksp = [Ag+]2 [SO42-]

Therefore, [SO42-] = Ksp / [Ag+]2

= (1.1 x 10-5) / (2.8 x 10-2)2

= 0.1403 x 10-1

= 1.403 x 10-2

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