Question

If 75.0 mL of a 2.63 · 10-3 M NaOH is mixed with 125.0 mL of 1.80 · 10-3 M MgCl2, then calculate the reaction quotient and state if a precipitate will form? The Ksp of the expected precipitate is 1.2 · 10-11.

Answer 1

Answer:

$Q =9.143x10^8$

Yes, the precipitate of magnesium hydroxide will be formed.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2NaOH(aq)+MgCl_2(aq)\rightarrow 2NaCl(aq)+Mg(OH)_2(s)$

Whereas the precipitate is the magnesium hydroxide which is formed by:

$2OH^-(aq)+Mg^{2+}(aq)\rightleftharpoons Mg(OH)_2(s)\ \ \ ; K=\frac{1}{K_{sp}}$

Take into account that the solubility product is the inverse reaction. In such a way, the equilibrium is:

$\frac{1}{K_{sp}}=\frac{1}{[OH^-]^2[Mg^{2+}]}$

Thus, we know the concentration of OH⁻⁻ from the concentration of sodium hydroxide being the same (2.63x10⁻³M) since it is a strong base. Moreover, since magnesium chloride totally dissociates into magnesium and chloride ions the 1.80x10⁻³ M is also the concentration of magnesium ions.

Next, as the solutions mix, the final concentration of OH⁻⁻ is:

$[OH^-]=\frac{75.0mL*2.63x10^{-3}M}{75.0mL+125.0mL}=9.86x10^{-4}M$

And the final concentration of Mg²⁺ is:

$[Mg^{2+}]=\frac{125.0mL*1.80x10^{-3}M}{75.0mL+125.0mL}=1.125x10^{-3}M$

Finally, we compute the reaction quotient:

$Q=\frac{1}{[OH^-]^2[Mg^{2+}]}=\frac{1}{(9.86x10^{-4})^2*1.125x10^{-3}} =9.143x10^8$

But the equilibrium constant:

$K=\frac{1}{K_{sp}}=\frac{1}{1.2x10^{-11}}=8.333x10^{10}$

Therefore, since K>Q, we infer that the precipitate of magnesium hydroxide (consider the procedure) will be formed.

Regards.