Answer to Question #153006 in General Chemistry for Joy

Question

Will a B(OH)₂ precipitate form if 50.0 mL of 0.03 M B₃X₂ is mixed with 350.0 mL of pH 11.5
Mg(OH)₂ solution at 25⁰C? Ksp of B(OH)₂ is 2.1 x 10⁻¹³.

Answer 1

$B_3X_2 + 3Mg(OH)_2 \to 3B(OH)_2 + Mg_3X_2$

pH + pOH = 14

pOH = 14 - 11.5 = 2.5

pOH = -log(OH-) = 2.5

OH = 10^-2.5

Therefore, [OH-] = 3.16 × 10^-3M

1 molecule of $Mg(OH)_2$ has 2 molecules of $OH^-$. Therefore, the concentration of $Mg(OH)_2$ is 6.32 × 10^-3M

$B(OH)_{2(s)} ↔ B^{2+} (aq) + 2OH^-_{(aq)}$; Ksp = 2.1 × 10^-13

mol B²⁺ (aq) = (0.050 L sol)(0.03M) = 0.0015 mol B²⁺

[B²⁺] = 0.5 mol/0.400 L = 0.00375 M

mol OH^- (aq) = 0.350 L sol)(0.00632M) = 0.00221 mol OH^-

[Cl—] = 0.00221 mol/0.4000 L = 0.00553 M

Q = [B²⁺][OH^-]²

 = (0.00375)(0.00553)²

 = 1.065 × 10^-7

Q > K_sp

B(OH)₂ will precipitate from solution.