Question

What mass of NaCN must be added to 1L of 0.010MMg (NO₃)₂in order to produce the first trace of Mg(OH)₂?

Short answer

The value of the mass produced in the first trade 3.6mg.

Step-by-step solution

Define concentration of the solution

A solution concentration is a measure of the quantity of solute that has been dissolved in a given quantity of solvent or solution. One that contains a relatively high volume of dissolved solute is a concentrated solution

Finding the solution that dissociates

Let us find the mass of NaCN that must be added to 1L of 0.010MMg (NO₃)₂ in order to precipitateMg(OH)₂

0.010M ofMg(OH)₂dissociates completely 0.010M ofMg²⁺and 0.020M²of NO₃^-

The dissociates of NaCN

NaCN (aq)⇌ Na⁺ (aq) +CN^-(aq)

Calculating the reaction CN, H2O

The reaction of CN^- andH₂O

CN^-(aq) +H₂O(I)⇌HCN (aq) + OH^- (aq)

K_a of HCN is 4.9.10^-10

\({K_b}{\rm{\;of\;C}}{{\rm{N}}^-}{\rm{is\;}}{K_b}=\frac{{{K_w}}}{{{K_a}}}=\frac{{1\cdot {{10}^{-14}}}}{{4.9\cdot{{10}^{-10}}}}=2.04\cdot{10^{-5}}\)

The reaction between Mg2+and OH-

The reaction between Mg²⁺and OH^-

Mg²⁺(aq) + 2OH^-(aq) ⇌ Mg(OH)₂(aq)

Let us calculate the solution =\[{K_{sp}}{\rm{\;of\;Mg}}{({\rm{OH}})_2}{\rm{\;is\;}}8.9\cdot {10^{-12}}\]

\begin{aligned}{{K_{sp}}=\left[{M{g^{2+}}}\right]\cdot{{\left[{O{H^-}}\right]}^2}}\\{8.9\cdot{{10}^{-12}}=0.010\cdot{{\left[{O{H^-}}\right]}^2}}\\{\left[{O{H^-}}\right]=2.98\cdot{{10}^{-5}}{\rm{M}}}\end{aligned}

 The concentration of the solution  OH- and HCN

Calculating the concentration of [CN^-] at equilibrium

\begin{aligned}{{K_b}=\frac{{[HCN]\cdot\left[{O{H^-}}\right]}}{{\left[{C{N^-}}\right]}}}\\{\left[{C{N^-}}\right]=\frac{{[HCN]\cdot\left[{O{H^-}}\right]}}{{{K_b}}}}\\{=\frac{{2.98\cdot{{10}^{-5}}\cdot2.98\cdot{{10}^{-5}}}}{{2.04\cdot{{10}^{-5}}}}}\\{=4.36\cdot{{10}^{-5}}{\rm{M}}}\end{aligned}

The mass of NaCN needed

Since ofreacted withto formanddid not react with total amount needed is

\begin{aligned}{[NaCN]=\left[{C{N^-}}\right]+[HCN]}\\{=4.36\cdot{{10}^{-5}}{\rm{M}}+2.98\cdot{{10}^{-5}}{\rm{M}}}\\{=7.35\cdot{{10}^{-5}}{\rm{M}}}\end{aligned}

\begin{aligned}{{n_{NaCN}}=[NaCN]\cdot1{\rm{L}}}\\{=7.35\cdot{{10}^{-5}}{\rm{M}}\cdot1{\rm{L}}}\\{=7.35\cdot{{10}^{-5}}{\rm{mol}}}\\{{m_{NaCN}}={n_{NaCN}}\cdot{M_{NaCN}}}\\{=7.35\cdot{{10}^{-5}}{\rm{mol}}\cdot49.0072{\rm{g}}/{\rm{mol}}}\\{=3.6\cdot {{10}^{-3}}{\rm{g}}}\\{=3.6{\rm{mg}}}\end{aligned}