1. What is the molar solubility of barium fluoride, BaF

BaF

At 25

a) 7.1 x 10^{-4} |
b) 1.0 x 10^{-3} |
c) 6.3 x 10^{-3} |
d) 7.9 x 10^{-3} |
e) 1.0 x 10^{-2} |

a) 0.12 g / L | b) 0.18 g / L | c) 1.10 g / L | d) 1.39 g / L | e) 1.75 g / L |

NaF(s) ==========> Na

When the BaF

BaF

but there is already some F

a) 2.5 x 10^{-6} |
b) 5.0 x 10^{-6} |
c) 2.5 x 10^{-5} |
d) 5.0 x 10^{-3} |
e) 6.3 x 10^{-3} |

a) 4.4 x 10^{-4} |
b) 8.8 x 10^{-4} |
c) 4.4 x 10^{-3} |
d) 8.8 x 10^{-1} |
e) 1.1 x 10^{0} |

CdS(s) <----------> Cd

At 25

a) 6.4 x 10^{-53} |
b) 4.0 x 10^{-27} |
c) 8.0 x 10^{-27} |
d) 8.9 x 10^{-14} |
e) 4.4 x 10^{-14} |

CdS(s) <----------> Cd

However, if NH

Cd

The overall equilibrium reaction is the sum of the above two reactions. After finding this overall reaction, and the numerical value of its associated equilibrium constant, you can do an ICE table calculation to find the molar solubility. It will be necessary to make an approximation to simplify the equation, just as I have done in class. Without this approximation, the mathematics will be horrible!

a) 8.0 x 10^{-20} |
b) 4.0 x 10^{-20} |
c) 2.0 x 10^{-10} |
d) 2.8 x 10^{-10} |
e) 3.2 x 10^{3} |

AgC

Would a precipitate of AgC

a) Yes, it would | b) No, it would not |

PbCl

PbI

If a solution contained Cl

a) PbCl_{2} |
b) PbI_{2} |

a) 3.2 X 10^{-7} |
b) 1.0 x 10^{-6} |
c) 4.1 x 10^{-5} |
d) 2.0 x 10^{-3} |
e) 6.4 x 10^{-3} |

HCl(aq) ==========> H

a) 1.0 x 10^{-16} |
b) 1.0 x 10^{-14} |
c) 1.0 x 10^{-12} |
d) 1.0 x 10^{-7} |
e) 1.0 x 10^{-2} |

NH

a) 1.5 | b) 2.3 | c) 4.8 | d) 9.2 | e) 11.7 |

HC

a) 2.2 | b) 5.9 | c) 8.1 | d) 11.8 | e) 13.6 |

HCN(aq) <----------> H

a) 1.0 x 10^{-14} |
b) 5.7 x 10^{-10} |
c) 2.3 x 10^{-7} |
d) 1.8 x 10^{-5} |
e) 6.3 x 10^{-3} |

NaOH(s) ==========> Na

a) 1.6 | b) 2.5 | c) 11.5 | d) 12.4 | e) 13.8 |

KC

The K

C

The numerical value of the equilibrium constant for the above reaction is normally not published in tables, because it can be obtained from the equilibrium constants of two other reactions that ARE published:

HC

H

By combining these two reactions in the appropriate way, you can produce the reaction for which the numerical value of the equilibrium constant is desired. Once you have that, you can do an ICE table calculation in the usual way. It should be possible to make an approximation in your solution to the ICE table calculation to avoid having to solve the quadratic formula.

a) 2.8 | b) 4.4 | c) 7.5 | d) 8.4 | e) 9.6 |

NH

The Cl

NH

The numerical value of the equilibrium constant for the above reaction is normally not published in tables, because it can be obtained from the equilibrium constants of two other reactions that ARE published:

NH

H

By combining these two reactions in the appropriate way, you can produce the reaction for which the numerical value of the equilibrium constant is desired. Once you have that, you can do an ICE table calculation in the usual way. It should be possible to make an approximation in your solution to the ICE table calculation to avoid having to solve the quadratic formula.

a) 4.1 x 10^{-12} |
b) 3.8 x 10^{-10} |
c) 1.3 x 10^{-7} |
d) 2.6 x 10^{-5} |
e) 7.4 x 10^{-3} |

ANSWERS:

1 C 2 C 3 C 4 C

5 D 6 D 7 B 8 B

9 D 10 C 11 E 12 D

13 D 14 D 15 E 16 B