Answer to Question #201623 in General Chemistry for Mickayla

Question

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. 1.46 x 10^-12 M
2. 1.46 x 10^-9 M
3. 3.64 x 10^-10 M
4. 1.31 x 10^-9 M
5. 1.31 x 10^-12 M
6. 3.28 x 10^-10 M

Answer 1

Solution

Solubility PbF₂=0.0175w/v

Molar mass = 245.2g/mol

PbF_2(s)↔️ Pb²⁺_(aq)+2F^-_(aq)

k_sp=[Pb²⁺] [F^-]²

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

Dissociation equation shows for every mole of PbF₂ that dissociates, 1mol of Pb²⁺ and 2 mol of F^- are produced

Pb²⁺= 7.137$\times10^{-5}$

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

K_sp=(7.137$\times 10^{-5})(1.427\times10^{-4})$ ²

=1.4533$\times 10^-12$

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