Question

Solid NaI is slowly added to a solution that is \(0.010 M\) in \(\mathrm{Cu}^{+}\) and \(0.010 \mathrm{M}\) in \(\mathrm{Ag}^{+}\). (a) Which compound will begin to precipitate first? (b) Calculate \(\left[\mathrm{Ag}^{+}\right]\) when CuI just begins to precipitate. (c) What percentage of \(\mathrm{Ag}^{+}\) remains in solution at this point?

Short answer

(a) AgI will begin to precipitate first. (b) [Ag+] = \(8.3 \times 10^{-7} M\) when CuI begins to precipitate. (c) 8.3 x \(10^{-3}%\) of Ag+ remains in solution at this point.

Step-by-step solution

Identify the Solubility Product Constants (\(K_{sp}\))

From a reference table, the \(K_{sp}\) for AgI and CuI can be found. Let's suppose \(K_{sp} (AgI) = 8.3 \times 10^{-17}\) and \(K_{sp} (CuI) = 1.0 \times 10^{-12}\).

Determine the Ion Concentration Needed for Precipitation

To determine which compound will precipitate first, the concentration of \(I^-\) ions needed to begin precipitation of the respective cation can be calculated. To do this, rearrange the \(K_{sp}\) equation \(K_{sp} = [Cu^+][I^-]\) or \(K_{sp} = [Ag^+][I^-]\) to give \([I^-] = \frac{K_{sp}}{[M^+]}\), where \(M^+\) is the metal ion. For AgI, this gives \([I^-] = \frac{8.3 \times 10^{-17}}{0.01} = 8.3 \times 10^{-15} M\) and for CuI, this gives \([I^-] = \frac{1.0 \times 10^{-12}}{0.01} = 1.0 \times 10^{-10} M\).

Identify the Compound That Will Precipitate First

Comparing the calculated \([I^-]\) concentrations, CuI requires a lower concentration to begin precipitating (\(1.0 \times 10^{-10} M\) compared to \(8.3 \times 10^{-15} M\) for AgI). Hence, AgI will precipitate first as its decreased solubility will cause it to reach its saturation point before CuI.

Calculate the [Ag+] When CuI Begins to Precipitate

When CuI starts to precipitate, \([I^-]\) = \(1.0 \times 10^{-10} M\). Substituting this into the Ksp expression for AgI (\(K_{sp} = [Ag^+][I^-]\)) gives \([Ag^+] = \frac{K_{sp}}{[I^-]} = \frac{8.3 \times 10^{-17}}{1.0 \times 10^{-10}} = 8.3 \times 10^{-7} M\).

Calculate the Percentage of [Ag+] Remaining

The initial concentration of \([Ag^+]\) was \(0.01 M\). Thus, the percentage remaining when CuI starts to precipitate can be calculated as Percentage Remaining = \(\frac{Final [Ag^+]}{Initial [Ag^+]} \times 100% = \frac{8.3 \times 10^{-7}}{0.01} \times 100% = 8.3 \times 10^{-3}%\).

Key concepts

Precipitation Reactions

Precipitation reactions occur when two soluble substances react in solution to form an insoluble solid, known as a precipitate. This type of reaction is central to the study of aqueous chemistry and plays a vital role in various fields, from environmental science to medicine.

In a typical precipitation reaction, ions in solution combine to form a compound that is not soluble in water. This happens because the product of the ion concentrations exceeds the solubility product constant (\( K_{sp} \) for the compound. In simple terms, when the chemical 'soup' becomes too 'thick' with specific ions, a solid will form and fall out of the solution.

An illustration of this concept comes from the exercise where NaI is added to a solution containing \( \text{Cu}^+ \) and \( \text{Ag}^+ \) ions. The \( I^- \) ions from NaI will react with these metal ions to form either AgI or CuI. By comparing the \( K_{sp} \) values of AgI and CuI, we determined that AgI has a lower \( K_{sp} \) and will precipitate first. The exercise demonstrates the effectiveness of using \( K_{sp} \) and ion concentration calculations to predict the outcome of precipitation reactions.

Solubility Equilibrium

Solubility equilibrium refers to the dynamic balance that exists between a dissolved substance (the solute) and its undissolved excess (the precipitate). At this equilibrium point, the rate at which the solute dissolves in the solvent equals the rate at which it precipitates out. The ratio of these concentrations at equilibrium is represented by a constant value known as the solubility product constant (\( K_{sp} \).

The \( K_{sp} \) is unique for every compound at a given temperature and provides insight into the compound's solubility. Substances with high \( K_{sp} \) values are considered to be more soluble in water than those with low \( K_{sp} \) values. In the context of our exercise, we can see how solubility equilibrium plays a role. The \( K_{sp} \) of AgI is much lower than that of CuI, indicating that AgI is less soluble and will reach equilibrium — and start to precipitate — at a lower concentration of \( I^- \) ions as compared to CuI.

Chemical Concentration Calculations

Accurate chemical concentration calculations are fundamental to predicting and understanding reactions in solution, including precipitation. Such calculations involve determining the molar concentration (\( [M] \) of ions or molecules in a solution, typically expressed in moles per liter (M).

These calculations often require a basic understanding of stoichiometry and the use of the solubility product constant (\( K_{sp} \)). For instance, to calculate which compound, AgI or CuI, will precipitate first from a solution containing \( \text{Ag}^+ \) and \( \text{Cu}^+ \) ions, we rely on \( K_{sp} \) values and the initial metal ion concentrations. By rearranging the equilibrium expression \( K_{sp} = [Ag^+][I^-] \) or \( K_{sp} = [Cu^+][I^-] \) to solve for \( [I^-] \) against known \( K_{sp} \) values, we find the critical concentration at which each ionic compound will start to precipitate. This step-by-step process ultimately leads to the determination of the percentage of the original silver ion concentration remaining in solution when copper(I) iodide begins to precipitate.