Question #201623

The solubility of lead(II) fluoride in water is 0.0175% (w/v) at 29°C. Calculate the K_{sp} of this salt at this temperature

- 1.46 x 10
^{-12}M - 1.46 x 10
^{-9}M - 3.64 x 10
^{-10}M - 1.31 x 10
^{-9}M - 1.31 x 10
^{-12}M - 3.28 x 10
^{-10}M

Expert's answer

Solution

Solubility PbF_{2}=0.0175w/v

Molar mass = 245.2g/mol

PbF_{2(s)}↔️ Pb^{2+}_{(aq)}*+2F*^{-}_{(aq)}

*k*_{sp}*=[Pb*^{2+}*] [F*^{-}*]*^{2}

Molar solubility= "\\frac{0.0175}{L}\\times\\frac{1mol}{245.2}= 7.137\\times10^{-5}"

Dissociation equation shows for every mole of PbF_{2 } that dissociates, 1mol of Pb^{2+} and 2 mol of F^{-} are produced

Pb^{2+}= 7.137"\\times10^{-5}"

F^{-= }"2\\times7.137\\times10^{-5}\n=1.427\\times10^{-4}"

K_{sp}=(7.137"\\times 10^{-5})(1.427\\times10^{-4})" ^{2}

=1.4533"\\times 10^-12"

=1.46"\\times 10^{-12}M" M

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