Problem 73 - General Chemistry: Principles and Modern Applications

Question

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

Answer

The molar solubilities of \(MgCO_3\) can be found by evaluating the concentration of each ion at equilibrium for each solution. The specific results depend on the solubility product constant of \(MgCO_3\) and the concentrations of \(NH_4^{+}\) and \(NH_3\) in each solution.

Step 1: 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'.

Step 2: 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 \(s\).

Step 3: 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.

Step 4: 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.

Key Concepts

Solubility Product Constant (Ksp)

Understanding the solubility product constant, or Ksp, is key when dealing with sparingly soluble compounds like magnesium carbonate (\(MgCO_3\)). Ksp is a mathematical expression that represents the equilibrium between a solid substance and its dissolved ions in a solution. It is defined only for ionic compounds that dissolve to a very limited extent in water.

For any sparingly soluble ionic compound \(AB_s\), the dissolution can be represented as \(AB_s(s) \rightleftharpoons A^+ (aq) + B^- (aq)\). The solubility product \(K_{sp}\) is expressed as the product of the equilibrium ion concentrations raised to the power of their stoichiometric coefficients in the balanced dissolution equation. For our example \(MgCO_3\), the dissociation is \(MgCO_3(s) \rightleftharpoons Mg^{2+}(aq) + HCO_3^{-}(aq)\).

• The expression becomes: \(K_{sp} = [Mg^{2+}][HCO_3^{-}]\).
• This indicates that \(K_{sp}\) is crucial in determining how much \(MgCO_3\) can dissolve in a solution to reach equilibrium.
• The concentrations of the ions are very low at this level of dissolution, making \(K_{sp}\) particularly useful.

We calculate the molar solubility by determining how many moles of \(MgCO_3\) can dissolve per liter until the solution reaches its equilibrium. By comparing conditions with different initial ion concentrations, such as in the presence of other ions or buffer solutions, we can see how the solubility changes.

Common Ion Effect

The common ion effect is a phenomenon where the solubility of an ionic compound is reduced due to the addition of a common ion already present in the solution. This occurs according to Le Châtelier's principle, which describes how an equilibrium system shifts to accommodate changes.

In our example involving \(MgCO_3\), the addition of \(NH_4^{+}\) from \(NH_4Cl\) introduces a common ion because \(NH_4^+\) is already participating in the reaction. This reduces the solubility product \(K_{sp}\) as the system will favor the solid form of \(MgCO_3\) to counterbalance the disturbance, thereby preventing further dissolution.

• In a solution containing \(1.00M\) \(NH_4Cl\), the presence of \(NH_4^{+}\) limits the dissociation of \(MgCO_3\).
• The resultant expression – \(K_{sp} = s * (1 - s)\) – captures how the solubility \(s\) gets affected.
• This means only a modest amount of \(MgCO_3\) dissolves, as more \(MgCO_3\) solidifies in response to elevated \(NH_4^{+}\) levels.

Thus, the effect is practical in changing the conditions under which a salt can dissolve or precipitate in the presence of compounds that contain one of its ions.

Buffer Solutions

A buffer solution is a special type of solution that resists changes in its pH upon the addition of small amounts of acid or base. This solution typically consists of a weak acid and its conjugate base or a weak base and its conjugate acid. In our problem, the buffer contains both \(NH_3\), a weak base, and \(NH_4^+\), its conjugate acid.

In the presence of a buffer, the solubility of \(MgCO_3\) is systematically altered. The properties of a buffer allow it to stabilize the pH, which in turn affects the equilibrium position of the \(MgCO_3\) dissolution reaction.

• In a \(1.00M\) \(NH_3\) and \(1.00M\) \(NH_4Cl\) buffer, the pH is maintained by the buffer, aiding a stable environment for the dissolution process. This scenario shifts the equilibrium to the right, thus increasing \(MgCO_3\) solubility.
• Similarly, when the buffer composition changes to \(0.100M\) \(NH_3\) and \(1.00M\) \(NH_4Cl\), the reduction in \(NH_3\) concentration (means less conversion capacity) influences the dissolution dynamic.
• Equilibrium expressions take into account the buffering capacity and components, balancing the system naturally.

Buffers are incredibly useful in maintaining consistency in reactions sensitive to pH changes, ensuring controlled experimental conditions.