Question

at equilibrium, a 1.0 l reaction vessel contains 2.3 mol of mg(oh)₂ and 0.170 mol of oh⁻. what is the equilibrium concentration of mg²⁺ based on the reaction: mg(oh)₂(s) ⇌ mg²⁺(aq) + 2oh⁻(aq) kₑ = 1.80×10⁻¹¹ answer:

Explanation:

Step1: Determine concentrations

Since volume is 1.0 L, $[OH^-]=\frac{0.170\ mol}{1.0\ L}= 0.170\ M$. Solids are not included in $K_c$ expression, so we focus on the ions.

Step2: Write $K_c$ expression

For the reaction $\text{Mg(OH)}_2(\text{s})
ightleftharpoons\text{Mg}^{2 +}(\text{aq})+2\text{OH}^-(\text{aq})$, the $K_c$ expression is $K_c=[\text{Mg}^{2 +}][\text{OH}^-]^2$.

Step3: Solve for $[\text{Mg}^{2 +}]$

We know $K_c = 1.80\times10^{-11}$ and $[\text{OH}^-]=0.170\ M$. Rearranging the $K_c$ expression for $[\text{Mg}^{2 +}]$ gives $[\text{Mg}^{2 +}]=\frac{K_c}{[\text{OH}^-]^2}$. Substituting the values: $[\text{Mg}^{2 +}]=\frac{1.80\times 10^{-11}}{(0.170)^2}$.
$[\text{Mg}^{2 +}]=\frac{1.80\times 10^{-11}}{0.0289}\approx6.23\times 10^{-10}\ M$.

Answer:

$6.23\times 10^{-10}\ M$