Question

Potassium hydroxide is used to precipitate each of the cations from their respective solution. Determine the minimum concentration of KOH required for precipitation to begin in each case.

Part A

1.9×10⁻² M CaCl2

Express your answer using two significant figures.

Part B

2.8×10⁻³ M Fe(NO3)2

Express your answer using two significant figures.

Part C

2.0×10⁻³ M MgBr2

Express your answer using two significant figures.

Thanks

Answer 1

A) The balanced chemical equation for the reaction between CaCl₂ and NaOH is given as

CaCl₂ (aq) + 2 KOH (aq) ---------> Ca(OH)₂ (s) + 2 KCl (aq)

Ca(OH)₂ (s) <=======> Ca²⁺ (aq) + 2 OH^- (aq)

K_sp = [Ca²⁺][OH^-]² = 5.02*10^-6

(http://www2.chm.ulaval.ca/gecha/chm1903/6_solubilite_solides/solubility_products.pdf)

Therefore,

5.02*10^-6 = (1.9*10^-2)*[OH^-]²

(ignore units since K_sp is dimensionless)

=====> [OH^-]² = (5.02*10^-6)/(1.9*10^-2) = 2.642*10^-4

=====> [OH^-] = 0.0162 ≈ 0.016 (correct to 2 sig. figs)

The minimum concentration of KOH required to precipitate 1.9*10^-2 M CaCl₂ as Ca(OH)₂ is 0.016 M (ans).

B) The balanced chemical equation for the reaction between CaCl₂ and NaOH is given as

Fe(NO₃)₂ (aq) + 2 KOH (aq) ---------> Fe(OH)₂ (s) + 2 KNO₃ (aq)

Fe(OH)₂ (s) <=======> Fe²⁺ (aq) + 2 OH^- (aq)

K_sp = [Fe²⁺][OH^-]² = 4.87*10^-17

(http://www2.chm.ulaval.ca/gecha/chm1903/6_solubilite_solides/solubility_products.pdf)

Therefore,

4.87*10^-17 = (2.8*10^-3)*[OH^-]²

(ignore units since K_sp is dimensionless)

=====> [OH^-]² = (4.87*10^-17)/(2.8*10^-3) = 1.739*10^-14

=====> [OH^-] = 1.318*10^-7 ≈ 1.3*10^-7 (correct to 2 sig. figs)

The minimum concentration of KOH required to precipitate 2.8*10^-3 M Fe(NO₃)₂ as Fe(OH)₂ is 1.3*10^-7 M (ans).

C) The balanced chemical equation for the reaction between CaCl₂ and NaOH is given as

MgBr₂ (aq) + 2 KOH (aq) ---------> Mg(OH)₂ (s) + 2 KBr (aq)

Mg(OH)₂ (s) <=======> Mg²⁺ (aq) + 2 OH^- (aq)

K_sp = [Mg²⁺][OH^-]² = 5.61*10^-12

(http://www2.chm.ulaval.ca/gecha/chm1903/6_solubilite_solides/solubility_products.pdf)

Therefore,

5.61*10^-12 = (2.0*10^-3)*[OH^-]²

(ignore units since K_sp is dimensionless)

=====> [OH^-]² = (5.61*10^-12)/(2.0*10^-3) = 2.805*10^-9

=====> [OH^-] = 5.296*10^-5 ≈ 5.3*10^-5 (correct to 2 sig. figs)

The minimum concentration of KOH required to precipitate 2.0*10^-3 M MgBr₂ as Mg(OH)₂ is 5.3*10^-5 M (ans).