The dissolution of \(\mathrm{MgCO}_{3}(\mathrm{s})\) in \(\mathrm{NH}_{4}^{+}(\mathrm{aq})\) can be represented as \(\mathrm{MgCO}_{3}(\mathrm{s})+\mathrm{NH}_{4}^{+}(\mathrm{aq}) \rightleftharpoons\) $$ \mathrm{Mg}^{2+}(\mathrm{aq})+\mathrm{HCO}_{3}^{-}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) $$ Calculate the molar solubility of \(\mathrm{MgCO}_{3}\) in each of the following solutions: (a) \(1.00 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq}) ;\) (b) a buffer that is \(1.00 \mathrm{M} \mathrm{NH}_{3}\) and \(1.00 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl} ;\) (c) a buffer that is \(0.100 \mathrm{M} \mathrm{NH}_{3}\) and \(1.00 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}.\)

Step by step solution

01

Write the balanced equation for the dissolution of \(MgCO_3\)

The balanced equation for the dissolution of \(MgCO_3\) in \(NH_{4}^{+}\) is given as: \[MgCO_{3}(s) + NH_{4}^{+}(aq) \rightleftharpoons Mg^{2+}(aq) +HCO_{3}^{-}(aq) +NH_{3}(aq)\]. Each mole of \(MgCO_{3}\) that dissolves produces 1 mole of \(Mg^{2+}\), \(HCO_{3}^{-}\), and \(NH_{3}\). Let's designate the molar solubility of \(MgCO_{3}\) as 's'. At equilibrium, the concentration of these products would be 's', assuming no other reactions involving these species. The concentration of \(NH_{4}^{+}\) would be slightly reduced due to reaction with \(MgCO_{3}\), so we can call it '1-s'.

02

Calculate molar solubility in \(1.00 M NH_{4} Cl\)

\[\[MgCO_{3}(s) + NH_{4}^{+}(aq) \rightleftharpoons Mg^{2+}(aq) +HCO_{3}^{-}(aq) +NH_{3}(aq)\] For the \(MgCO_3\) dissolution process, under saturated solution conditions, the molar ion concentration relation can be written as: \[Ksp = [Mg^{2+}][HCO_3^{-}] = s^2\] The value of Ksp can be found in a standard table of solubility products. Given solution is \(1.00M\) of \(NH_4 Cl\), which means the product of the solubility from the reaction and \(NH_4^{+}\) is 1. Since \(NH_4^{+}\) is common ion, its concentration remains virtually unchanged: \[Ksp = s * (1-s)\] When solved, we get the molar solubility \(s\).

03

Calculate molar solubility in \(1.00M NH_3\) and \(1.00M NH_4 Cl\) buffer

Buffers that contain \(NH_3\) will shift the equilibrium of dissolution of \(MgCO_3\) to the right because of the reduction in the \(NH_3\) concentration produced by the original reaction. This will result in an increased solubility of \(MgCO_3\). We can denote the amount of \(NH_3\) consumed by the reaction to be 'x'. So, the \(NH_3\) concentration becomes '1-x' while the concentration of \(NH_4^{+}\) becomes '1+x'. The solubility product expression now becomes: \[Ksp = [Mg^{2+}][HCO_3^{-}]*(1-x)\]. Solving this equation gives the molar solubility in the buffer.

04

Calculate molar solubility in \(0.100M NH_3\) and \(1.00M NH_4 Cl\) buffer

This is similar to the procedure in step 3 but with a reduced concentration of \(NH_3\) in the buffer. So, the \(NH_3\) concentration becomes '0.100-x' while the concentration of \(NH_4^{+}\) becomes '1+x'. The solubility product expression now becomes: \[Ksp = [Mg^{2+}][HCO_3^{-}]*(0.100-x)\]. Solution of this equation gives the molar solubility in this buffer.

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