Question

The concentration of fluoride ions in a saturated solution of barium fluoride is ________ m. the solubility product constant of baf2 is 1.7 ⋅ 10-6.

Answer 1

Answer:

The concentration of fluoride ions in a saturated solution of barium fluoride is 0.001302 M.

Explanation:

$BaF_2\rightleftharpoons Ba^{2+}+2F^-$

                S         2S

Solubility product of barium fluoride =$K_{sp}=1.7\times 10^{-6}$

The expression of solubility product is given as:

$K_{sp}=[Ba^{2+}][F^-]^2=[S][2S]^2=4S^2$

$1.7\times 10^{-6}=4S^2$

$S=6.51\times 10^{-4} M$

The concentration of fluoride ions =2S =$2\times 6.51\times 10^{-4} M=0.001302 M$

The concentration of fluoride ions in a saturated solution of barium fluoride is 0.001302 M.

Answer 2

The concentration of fluoride ions, in the solution is 1.504×10¯² M

Dissociation equation

BaF₂(aq) <=> Ba²⁺(aq) + 2F¯(aq)

Let the concentration of Ba²⁺ be y.

Then,

The concentration of F¯ will be 2y

Determination of the value of y

•Solubility of product (Kₛₚ) = 1.7×10¯⁶

•Value of y =?

Kₛₚ = [Ba²⁺][F¯]²

1.7×10¯⁶ = y × (2y)²

1.7×10¯⁶ = 4y³

Divide both side by 4

y³ = 1.7×10¯⁶ / 4

Take the cube root of both side

y = ³√(1.7×10¯⁶ / 4)

y = 7.52×10¯³ M

Determination of concentration of the fluoride ion, F¯

•y = 7.52×10¯³ M

• Concentration of Fluoride ion [F¯] =?

[F¯] = 2y

[F¯] = 2 × 7.52×10¯³

Concentration of Fluoride ion [F¯] = 1.504×10¯² M

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