Question

In the qualitative analysis scheme, magnesium and nickel precipitate from solution upon the addition of sodium hydroxide. Once separated from the remaining cations by filtration or decanting, the solid mixture is acidified and warmed to dissolve magnesium and nickel cations back into solution. Adding ammonia creates a buffer solution. (Remember, HCl and NH3 makes for NH4^+ cation.) The buffered solution should just be basic (say, pH = 8). Adding sodium hydrogen phosphate (Na​2HPO4) precipitates magnesium as MgNH4PO4 (Ksp = 3 x 10^-13).
Ksp = [Mg^2+​]*[NH3]*[HPO4^2-​] = 3 x 10^-13
Molar solubility = 6.7 x 10^-5 M
a) Calculate the [NH4^+] in a pH of 8.
b) Calculate the molar solubility of MgNH4PO4 in a solution with pH of 8.

Answer 1

Answer on Question #63480 - Chemistry - General Chemistry

In the qualitative analysis scheme, magnesium and nickel precipitate from solution upon the addition of sodium hydroxide. Once separated from the remaining cations by filtration or decanting, the solid mixture is acidified and warmed to dissolve magnesium and nickel cations back into solution. Adding ammonia creates a buffer solution. (Remember, HCl and $\mathrm{NH}_3$ makes for $\mathrm{NH}_4^+$ cation.) The buffered solution should just be basic (say, pH = 8). Adding sodium hydrogen phosphate (Na₂HPO₄) precipitates magnesium as MgNH₄PO₄ (Ksp = 3×10⁻¹³).

$$K_{\mathrm{sp}} = [\mathrm{Mg}^{2+}]^* [\mathrm{NH}_3]^* [\mathrm{HPO}_4^{2-}] = 3 \times 10^{-13}.$$

Molar solubility = 6.7 × 10⁻⁵ M

a) Calculate the $[\mathrm{NH}_4^+]$ in a pH of 8.

b) Calculate the molar solubility of $\mathrm{MgNH_4PO_4}$ in a solution with pH of 8.

Solution.

$$\mathrm{Mg}^{2+} + 2\mathrm{NaOH} \rightarrow \mathrm{Mg(OH)}_2\downarrow + 2\mathrm{Na}^+$$ $$\mathrm{Mg(OH)}_2 + 2\mathrm{H}^+ \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{H}_2\mathrm{O}$$ $$\mathrm{Mg}^{2+} + \mathrm{HPO}_4^{2-} + \mathrm{NH}_3 \rightarrow \mathrm{MgNH}_4\mathrm{PO}_4\downarrow$$

a) $\mathrm{pH} = 8$

$$\mathrm{NH}_3 + \mathrm{HCl} = \mathrm{NH}_4\mathrm{Cl}$$

V (solution) = 1 L; n₀(NH₃) = 1 mol; C₀(NH₃) = 1 mol/L

$$C(\mathrm{NH}_4^+) = x \ \mathrm{mol/L}; \quad C(\mathrm{NH}_3) = (1 - x) \ \mathrm{mol/L}$$ $$pH = pK_{a,\mathrm{NH}_3^+} - \lg \frac{C_{\mathrm{NH}_3^+}}{C_{\mathrm{NH}_3}}$$ $$K_{b,\mathrm{NH}_3} = 1.76 \cdot 10^{-5}; \quad pK_{b,\mathrm{NH}_3} = -\lg K_{b,\mathrm{NH}_3} = -\lg (1.76 \cdot 10^{-5}) = 4.75;$$ $$pK_{a,\mathrm{NH}_4^+} = pK_w - pK_{b,\mathrm{NH}_3} = 14 - 4.75 = 9.25$$ $$pH = 9.25 - \lg \frac{x}{1 - x} = 8$$ $$x = (1 - x) \times 17.78$$ $$x = 17.78 - 17.78x$$ $$18.78x = 17.78$$ $$x = 0.95 \ \mathrm{mol/L}$$ $$[\mathrm{NH}_4^+] = 0.95 \ \mathrm{mol/L}$$

b) $K_{\mathrm{sp}} = [\mathrm{Mg}^{2+}]^* [\mathrm{NH}_4^+]^* [\mathrm{PO}_4^{3-}] = 3 \times 10^{-13}$

$$K_{\mathrm{sp}} = 5 \times 0.95 \times 5 = 3 \times 10^{-13}$$ $$S = \sqrt{\frac{K_{\mathrm{sp}}}{0.95}} = \sqrt{\frac{3 \times 10^{-13}}{0.95}} = 5.6 \times 10^{-7} \ M$$

Answer: a) $[\mathrm{NH}_4^+] = 0.95 \ \mathrm{mol/L}$

b) $S = 5.6 \times 10^{-7} \ M$

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