Answer to Question #237098 in General Chemistry for Nana

Question

When SrF₂, strontium fluoride, is added to water, the salt dissolves to a very small extent according to the reaction below. At equilibrium the concentration of Sr²⁺ is found to be 0.00196 M. What is the value of K_sp for SrF₂? (Please give your answer with 2 significant figures.)

Answer 1

$[Sr^{2+}] = 0.00196 \;M = S$

SrF₂ <=> Sr²⁺(aq) + 2F^-(aq)

$k_{sp} = [Sr^{+2}][F^-]^2 \\ k_{sp} = (0.00196)(2 \times 0.00196)^2 \\ k_{sp} = 3.0 \times 10^{-8}$