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

`Q =9.143x10^8`

Yes, the precipitate of magnesium hydroxide will be formed.

Step-by-step 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=(1)/(K_(sp))`

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

`(1)/(K_(sp))=(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^-]=(75.0mL*2.63x10^(-3)M)/(75.0mL+125.0mL)=9.86x10^(-4)M`

And the final concentration of Mg²⁺ is:

`[Mg^(2+)]=(125.0mL*1.80x10^(-3)M)/(75.0mL+125.0mL)=1.125x10^(-3)M`

Finally, we compute the reaction quotient:

`Q=(1)/([OH^-]^2[Mg^(2+)])=(1)/((9.86x10^(-4))^2*1.125x10^(-3)) =9.143x10^8`

But the equilibrium constant:

`K=(1)/(K_(sp))=(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.