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Untitled - Kelly Walsh High School

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Buffers and Other Equilibria 239<br />

Don’t Forget!<br />

There is never a denominator in a K sp expression.<br />

For this particular salt, the value of the K sp is 1.6 10 8 at 25C. If we know the<br />

value of the solubility product constant, then we can determine the concentration<br />

of the ions. In addition, if we know one of the ion concentrations, then we<br />

can find the K sp.<br />

The K sp of magnesium fluoride in water is 8 10 8 . How many grams of magnesium<br />

fluoride will dissolve in 0.250 L of water?<br />

MgF2(s) K Mg2 (aq) 2 F (aq)<br />

Ksp [Mg2 ][F ] 2 8 108 For every mole of MgF 2 that dissolves, 1 mol of Mg 2+ and 2 mol of F form:<br />

K sp (x)(2x) 2 4 x 3 8 10 8<br />

x 3 10 3 [Mg 2 ]<br />

a 0.05 g MgF2 3 103 mol Mg2 b (0.250 L) a<br />

L<br />

1 mol MgF2 1 mol Mg2b a62.3 g MgF2 b<br />

1 mol MgF2 When solving common-ion-effect problems, calculations like the ones above<br />

involving finding concentrations and Ksp’s can still be done, but the concentration<br />

of the additional common ion will have to be inserted into the solubility<br />

product constant expression. Sometimes, if the Ksp is very small and the common<br />

ion concentration is large, we can simply approximate the concentration of<br />

the common ion by the concentration of the ion added.<br />

Calculate the silver ion concentration in each of the following solutions:<br />

a. Ag2CrO4(s) water<br />

b. Ag2CrO4(s) 1.00 M Na2CrO4 Ksp 1.9 1012 a. Ag2CrO4(s) K 2 Ag 2 (aq) CrO4 (aq)<br />

2 x x<br />

Ksp 1.9 1012 (2 x) 2 (x) 4 x3 x 7.8 105 [Ag ] 2 x 1.6 104 M

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