Question

A volume of \(50mL\) of \(1.8MN{H_3}\)is mixed with an equal volume of a solution containing\(0.95g\;of\;MgC{l_2}\). What mass of \(N{H_4}Cl\)must be added to the resulting solution to prevent the precipitation of \(Mg{(OH)_2}?\)

Short answer

The volume of solution is \(9.18{\rm{g}}\)

Step-by-step solution

Define solution containing

The substance dissolved in the solution is called the solute, whereas the component in which the solute is dissolved is known as a solvent. The solution containing water as the solvent is called an aqueous solution

Calculating the concentration of Ammonia

\(50{\rm{mL\;of\;}}1.8{\rm{MN}}{{\rm{H}}_3}\)

\(50{\rm{mL}}\)of a solution with\(0.95{\rm{g\;of\;MgC}}{{\rm{l}}_2}\)

The total volume of a mixture is\(100{\rm{mL}}(0.100{\rm{L}})\)

The concentration of\({\rm{N}}{{\rm{H}}_3}\)is

\(\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = \frac{{1.8{\rm{M}} \cdot 50{\rm{mL}}}}{{100mL}} = 0.90{\rm{M}}\)

Concentration of \(MgC{l_2}\)

\(\left[ {Mg{\rm{C}}{{\rm{l}}_2}} \right] = \left[ {{\rm{M}}{{\rm{g}}^{2 + }}} \right] = \frac{{\frac{{0.95{\rm{g}}}}{{95.211{\rm{g}}/{\rm{mol}}}}}}{{0.100{\rm{L}}}} = 0.0998{\rm{M}}\)

Calculate the mass of\({\rm{N}}{{\rm{H}}_4}{\rm{Cl}}\)that must be added to prevent precipitation of\({\rm{Mg}}{({\rm{OH}})_2}\)

\({K_{sp}}{\rm{\;for\;Mg}}{({\rm{OH}})_2}{\rm{\;is\;}}8.9 \cdot {10^{ - 12}}\)

Calculate the concentration of\({\rm{O}}{{\rm{H}}^ - }\)ions in a saturated solution

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {{\rm{M}}{{\rm{g}}^{2 + }}} \right] \cdot {{\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}^2}}\\{8.9 \cdot {{10}^{ - 12}} = 0.0998 \cdot {{\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}^2}}\\{\left[ {O{H^ - }} \right] = 9.44 \cdot {{10}^{ - 6}}{\rm{M}}}\end{array}\)

Calculating the concentration of\(N{H_4} + \)

\({K_b}{\rm{\;of\;N}}{{\rm{H}}_3}{\rm{\;is\;}}{K_b} = 1.8 \cdot {10^{ - 5}}\)

\(\begin{array}{*{20}{c}}{{K_b} = \frac{{\left[ {{\rm{NH}}_4^ + } \right] \cdot \left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}}{{\left[ {N{H_3}} \right]}}}\\{1.8 \cdot {{10}^{ - 5}} = \frac{{\left[ {NH_4^ + } \right] \cdot 9.44 \cdot {{10}^{ - 6}}}}{{0.90}}}\\{\left[ {NH_4^ + } \right] = \frac{{1.8 \cdot {{10}^{ - 5}} \cdot 0.90}}{{9.44 \cdot {{10}^{ - 6}}}}}\\{ = 1.716{\rm{M}}}\end{array}\)

The concentration of \({\rm{N}}{{\rm{H}}_4}{\rm{Cl}}\)that must be added is\(\left[ {{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}} \right] = {\rm{N}}{{\rm{H}}_4} + = 1.716{\rm{M}}\)

Calculating the mass precipitation

Number of moles of\({\rm{N}}{{\rm{H}}_4}{\rm{Cl}}\)is

\(\begin{array}{*{20}{c}}{{n_{N{H_4}Cl}} = 1.716{\rm{M}} \cdot 0.100{\rm{L}}}\\{ = 17.16 \cdot {{10}^{ - 2}}{\rm{mol}}}\end{array}\)

The mass of\({\rm{N}}{{\rm{H}}_4}{\rm{Cl}}\)that must be added to prevent precipitation of\({\rm{Mg}}{({\rm{OH}})_2}\)is

\({m_{N{H_4}C}} = 17.16 \cdot {10^{ - 2}}{\rm{mol}} \cdot 53.491{\rm{g}}/{\rm{mol}}\)