### Calculate K_{sp} given a weight/volume percentage

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**Example #1:** The solubility of lead(II) fluoride is 0.0510% (w/v) at 25 °C. Calculate the K_{sp} of this salt at this temperature.

**Solution:**

1) 0.0510% (w/v) means:

0.0510 grams of PbF_{2} in 100. mL of solution.

2) Determine moles of PbF_{2} present (the molecular weight of PbF_{2} is 245.196 g mol¯^{1}):

0.0510 g / 245.196 g mol¯^{1} = 2.08 x 10¯^{4} mol

3) Determne the molarity of PbF_{2}:
2.08 x 10¯^{4} mol divided by 0.100 L = 2.08 x 10¯^{3} mol/L.

4) Determine K_{sp} of PbF_{2}:
PbF_{2} ⇌ Pb^{2+} + 2F¯
K_{sp} = [Pb^{2+}] [F¯]^{2}

K_{sp} = (2.08 x 10¯^{3}) (4.16 x 10¯^{3})^{2}

K_{sp} = 3.60 x 10¯^{10}

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