Question

Use the Solver function in Excel to determine [Sr2 ], [F–], [HF], [OH–], and [H ] for a saturated solution of SrF2. The pKsp for SrF2 is 8.58. The pKa for HF is 3.17. Ignore activity coefficients.

Answer 1

we know that Pka is 3.17

Ka = 10^-3.17 = 0.00067608

Kw = ka * kb, kw is dissociation constant for water which is 1x10^-14

so kb = kw / ka = 1x10^-14 / 0.00067608= 1.48 x 10^-11,

lets start analyzing the SFr₂ compound

pksp = 8.58

ksp = 10^-8.58 = 2.63 x 10^-9

so:

SFr₂ === S⁺² + 2 Fr^-

..................x.........2x.

ksp = [x] * [2x]² = 2.63 x 10^-9

4x³ = 2.63 x 10^-9

solve for x to get: 0.00087 M

so Sr⁺² = 0.00087 M

get the concentration of F

[F] = 2 * 0.00087 = 0.00174 M

F reacts with water to produce

F + H₂O === HF + OH^-

Kb = [OH] [ HF ] / F, we know that kb is 1.48 x 10^-11 this is a very small value so:

1.48 x 10^-11 = [OH] [ HF ] / F ,

1.48 x 10^-11 = [OH] [ HF ] /  0.00174

1.48 x 10^-11 = x² /  0.00174, solve for x

x = 1.6 x 10^-7 M = [HF ] = [OH]

[H] = 1 x 10^-14 / 1.6 x 10^-7 M = 6.23 x 10^-8 M

*hope it helps =)