Problem 89P

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

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

SrCO3 |
2.4 × 10−5 |
2.4 × 10−5 |
_____ |

SrF2 |
1.0 × 10−3 |
_____ |
4.0 × 10−9 |

Ag2CO3 |
_____ |
1.3 × 10−4 |
8.8 × 10−12 |

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 |

SrCO3 |
2.4 x 10-5 |
2.4 x 10-5 |
5.76 x 10-10 |

SrF2 |
1.0 x 10-3 |
2.0 x 10-3 |
4.0 x 10-9 |

Ag2CO3 |
2.6 x 10-4 |
1.3 x 10-4 |
8.8 x 10-12 |

- The ionization of SrCO3 takes place in the following manner:

SrCO3 Sr2+ + CO32-

Solubility product for the reaction, Ksp = [Sr2+][CO32-]

= (2.4 x 10-5)(2.4 x 10-5)

= 5.76 x 10-10

b) The ionization of SrF2 takes place in the following manner:

SrF2 Sr2+ + 2F-

Solubility product for the reaction, Ksp = [Sr2+][F-]2

Therefore, [F-]2 = Ksp / [Sr2+]

= (4.0 x 10-9) / (1.0 x 10-3)

= 4.0 x 10-6

[F-] =

= 2.0 x 10-3

c) The ionization of Ag2CO3 takes place in the following manner:

Ag2CO3 2Ag+ + CO32-

Solubility product for the reaction, Ksp = [Ag+]2 [CO32-]

Therefore, [Ag+]2 = Ksp / [CO32-]

= (8.8 x 10-12) / (1.3 x 10-4)

= 6.769 x 10-8

[Ag+] =

= 2.6 x 10-4

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