Question

determine the value of ksp for mg(cn)₂ by constructing an ice table, writing the solubility constant expression, and solving the expression. the molar solubility of mg(cn)₂ is 1.4 × 10⁻⁶ m at a certain temperature. complete parts 1 - 2 before submitting your answer. using the values from the ice table, construct the expression for the solubility constant. each reaction participant must be represented by one tile. do not combine terms once the expression is constructed, solve for ksp. ksp = =

Explanation:

Step1: Write the dissolution equation

\(Mg(CN)_2(s)
ightleftharpoons Mg^{2 +}(aq)+2CN^-(aq)\)
Let the molar solubility of \(Mg(CN)_2\) be \(s\). From the ICE table (not shown here but based on stoichiometry), \([Mg^{2+}]=s = 1.4\times10^{-6}M\) and \([CN^-]=2s=2.8\times 10^{-6}M\)

Step2: Construct the \(K_{sp}\) expression

The solubility - product constant expression for \(Mg(CN)_2\) is \(K_{sp}=[Mg^{2+}][CN^-]^2\)

Substitute \(s = 1.4\times10^{-6}\) into the expression:
\(K_{sp}=(1.4\times 10^{-6})\times(2.8\times 10^{-6})^2\)

Step3: Calculate \(K_{sp}\)

First, calculate \((2.8\times 10^{-6})^2=2.8^2\times10^{-12}=7.84\times 10^{-12}\)
Then, \(K_{sp}=(1.4\times10^{-6})\times(7.84\times 10^{-12})\)
\(K_{sp}=1.4\times7.84\times10^{-6 - 12}\)
\(K_{sp}=10.976\times10^{-18}=1.1\times 10^{-17}\)

Answer:

\(K_{sp}=\frac{[1.4\times 10^{-6}][2.8\times 10^{-6}]^2}{[0]} = 1.1\times 10^{-17}\)