Question

In qualitative analysis, \(\mathrm{Ca}^{2+}\) and \(\mathrm{Ba}^{2+}\) are separated from \(\mathrm{Na}^{+}, \mathrm{K}^{+}\), and \(\mathrm{Mg}^{2+}\) by adding aqueous \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) to a solution that also contains aqueous \(\mathrm{NH}_{3}\) (Figure 16.18). Assume that the concentrations after mixing are \(0.080 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) and \(0.16 \mathrm{M} \mathrm{NH}_{3}\). (a) List all the Brønsted-Lowry acids and bases present initially, and identify the principal reaction. (b) Calculate the \(\mathrm{pH}\) and the concentrations of all species present in the solution. (c) In order for the human eye to detect the appearance of a precipitate, a very large number of ions must come together to form solid particles. For this and other reasons, the ion product must often exceed \(K_{\mathrm{sp}}\) by a factor of about \(10^{3}\) before a precipitate can be detected in a typical qualitative analysis experiment. Taking this fact into account, show quantitatively that the \(\mathrm{CO}_{3}{ }^{2-}\) concentration is large enough to give observable precipitation of \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3}\), but not \(\mathrm{MgCO}_{3}\). Assume that the metal-ion concentrations are \(0.010 \mathrm{M}\). (d) Show quantitatively which of the \(\mathrm{Mg}^{2+}, \mathrm{Ca}^{2+}\), and \(\mathrm{Ba}^{2+}\) ions, if any, should give an observable precipitate of the metal hydroxide. (e) Could the separation of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Ba}^{2+}\) from \(\mathrm{Mg}^{2+}\) be accomplished using \(0.80 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) in place of \(0.080 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} ?\) Show quantitatively why or why not.

Short answer

Separation is feasible; \\mathrm{CaCO}_3\\ and \\mathrm{BaCO}_3\\ can precipitate, but not \\mathrm{MgCO}_3\\. \\mathrm{Na}_2\\mathrm{CO}_3\\ in higher concentration may enhance separation.

Step-by-step solution

Identify the Brønsted-Lowry Acids and Bases

Initially, the solution contains the following species: - \(\mathrm{NH}_3\), a weak base that can accept a proton. - \(\mathrm{NH}_4^+\), the conjugate acid of \(\mathrm{NH}_3\), which can donate a proton. - \(\mathrm{CO}_3^{2-}\) from \(\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3\), which can accept protons to form \(\mathrm{HCO}_3^-\) and ultimately \(\mathrm{H}_2\mathrm{CO}_3\) (water and \(\mathrm{CO}_2\)).The principal reaction that governs the pH of the solution is the equilibria involving \(\mathrm{NH}_3\) accepting a proton to form \(\mathrm{NH}_4^+\).

Calculate the pH of the Solution

To find the pH, consider the \(\mathrm{NH}_3\) and \(\mathrm{NH}_4^+\) equilibrium:\[\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^- \\mathrm{K_b} = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]}\]The \(K_b\) for \(\mathrm{NH}_3\) is approximately \(1.8 \times 10^{-5}\). Using the initial concentration and an ICE table, solve for \(\mathrm{OH}^-\) concentration, and convert to pH:1. Set up the equation with \(x\) as the change in concentration: \(0.16 - x\) for \(\mathrm{NH}_3\) and \(x\) for \(\mathrm{OH}^-\).2. Solve \(x^2 = K_b(0.16 - x)\), simplifying to find \(x\).3. Convert \(x\) (\(\mathrm{OH}^-\) concentration) to \(\mathrm{pH}\) using \(\mathrm{pOH} = -\log[x]\) and \(\mathrm{pH} + \mathrm{pOH} = 14\).

Determine \\(\mathrm{CO}_3^{2-}\\) Concentration and Precipitation Condition

Using the carbonate equilibrium and given conditions:- With \(\mathrm{NH}_3\) at \(0.16\) M and \((\mathrm{NH}_4)_2 \mathrm{CO}_3\) at \(0.080\) M, calculate \(\mathrm{CO}_3^{2-}\).- Assume \(\mathrm{CO}_3^{2-}\) partially dissociates, using the solubility product constants \(K_{sp}\) for calcium and barium carbonate precipitations.- \(K_{sp}\) values: \(K_{sp, CaCO_3} = 4.5 \times 10^{-9}\) and \(K_{sp, BaCO_3} = 5.1 \times 10^{-9}\).- Calculate ion products: \([\mathrm{Ca}^{2+}][\mathrm{CO}_3^{2-}]\) and \([\mathrm{Ba}^{2+}][\mathrm{CO}_3^{2-}]\) must exceed \(10^3 \times K_{sp}\).- Determine which forms a precipitate using: \(0.010 \mathrm{M}\) metal ion concentration and \(\mathrm{CO}_3^{2-}\) calculated concentration.

Precipitation of Metal Hydroxides

Check the hydroxide precipitation condition by examining whether any metal ions form insoluble hydroxides:- For hydroxides \(\text{K}_{sp, \text{M(OH)}_2}:\) calcium, barium, and magnesium hydroxide solubility products are necessary.- Compare \(\mathrm{OH}^-\) concentration from pH calculation to hydroxide \(K_{sp}\) requirements.- Use \(\text{K}_{sp, \text{Ca(OH)}_2} = 5.5 \times 10^{-6},\ \text{K}_{sp, \text{Ba(OH)}_2} = 5.0 \times 10^{-3}\) for assessment with given metal ion concentrations. Estimate if any metal hydroxide concentration exceeds \(K_{sp}\) for observable precipitation.

Evaluate Alternative Precipitation using Na2CO3

Evaluate \(0.80 \mathrm{M} \mathrm{Na}_2 \mathrm{CO}_3\) as the carbonate source:- Assume previous logic: evaluate using change primarily in carbonate source concentration.- Calculate new ~ Using \(0.80 \mathrm{M}\), re-calculate precipitate formation and metal ion solubility.- Use ion concentrations and solubility products, as done for \(0.080 \(\mathrm{NH}_4\)_2 \mathrm{CO}_3\).- Determine possible separation effectiveness for metal carbonates.- Analysis should show if enhanced product exceeding substantially helps exceed \(K_{sp}\) of calcium and barium carbonate.

Key concepts

Brønsted-Lowry Acids and Bases

In aqueous solutions, the Brønsted-Lowry theory helps identify substances that can donate or accept protons. According to this theory:

• Acids are proton donors.
• Bases are proton acceptors.

For the exercise, the main players are:

• \( \mathrm{NH}_3 \), a weak base that can accept protons to form \( \mathrm{NH}_4^+ \).
• \( \mathrm{NH}_4^+ \), the conjugate acid, which can donate protons.
• \( \mathrm{CO}_3^{2-} \), which can accept protons to form \( \mathrm{HCO}_3^- \) in a series of reactions that eventually lead to carbonic acid (\( \mathrm{H}_2\mathrm{CO}_3 \)).

The main reaction is the equilibrium between \( \mathrm{NH}_3 \) and \( \mathrm{NH}_4^+ \), which dictates the pH of the solution.

Precipitation Reactions

Precipitation occurs when ions in a solution form an insoluble compound, creating a solid. In this context, calcium and barium ion reactions with carbonate ions lead to the formation of insoluble carbonates:

• \( \mathrm{CaCO}_3 \)
• \( \mathrm{BaCO}_3 \)

For a precipitate to form visibly, the ion product of the reacting ions must exceed their solubility product constant \( K_{sp} \). In practical terms, this often means the ion product must be about \( 10^3 \) times the \( K_{sp} \) value. Therefore, calculating whether the concentrations of \( \mathrm{Ca}^{2+} \) and \( \mathrm{Ba}^{2+} \) with \( \mathrm{CO}_3^{2-} \) exceed this threshold determines precipitation.

pH Calculation

Calculating the pH involves understanding the equilibrium between ammonia \( \mathrm{NH}_3 \) and ammonium \( \mathrm{NH}_4^+ \). This equilibrium is essential in establishing the hydroxide concentration.
Using the relationship:\[\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^- \]The base ionization constant \( K_b \) for \( \mathrm{NH}_3 \) is \( 1.8 \times 10^{-5} \).
Here's how to calculate the pH:

• Use an ICE table to track changes in concentrations.
• Express \( K_b \) in terms of \( x \) (change in concentration).
• Solve for \( x \) (concentration of \( \mathrm{OH}^- \)).

Convert \( \mathrm{OH}^- \) concentration to pOH using:\[\mathrm{pOH} = -\log[\mathrm{OH}^-]\]Finally, use the relation \( \mathrm{pH} + \mathrm{pOH} = 14 \) to find the pH.

Solubility Product Constant

The solubility product constant \( K_{sp} \) is a crucial parameter in predicting whether a precipitate will form in ionically strong solutions.
Each ionic compound has a specific \( K_{sp} \) value.

• \( K_{sp, CaCO_3} = 4.5 \times 10^{-9} \)
• \( K_{sp, BaCO_3} = 5.1 \times 10^{-9} \)

To determine precipitation:

• Calculate the ion product \([\mathrm{M}^{2+}][\mathrm{CO}_3^{2-}]\).
• If this value exceeds \( 10^3 \times K_{sp} \), a precipitate is likely to form.

This method helps assess visible precipitation, as more ions must combine to make it observable.

Metal Hydroxides

Certain metal ions can form metal hydroxides as another type of precipitate under basic conditions.
The solubility product \( K_{sp} \) determines whether a metal hydroxide like \( \mathrm{M(OH)}_2 \) will form:

• \( K_{sp, \mathrm{Ca(OH)}_2} = 5.5 \times 10^{-6} \)
• \( K_{sp, \mathrm{Ba(OH)}_2} = 5.0 \times 10^{-3} \)

Use the hydroxide concentration from the pH calculation to evaluate precipitation.
The metal ion concentration must exceed the hydroxide's \( K_{sp} \) to precipitate.