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A solution contains \(2.0 \times 10^{-4} \mathrm{M} \mathrm{Ag}^{+}\) and \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Pb}^{2+}\). If NaI is added, will AgI \(\left(K_{s p}=8.3 \times 10^{-17}\right)\) or \(\mathrm{PbI}_{2}\left(K_{s p}=7.9 \times 10^{-9}\right)\) precipitate first? Specify the concentration of \(\mathrm{I}^{-}\) needed to begin precipitation.

Short Answer

Expert verified
AgI will precipitate first, and the minimum concentration of \(I^{-}\) needed for precipitation is \(4.15 \times 10^{-13} M\).

Step by step solution

01

Write the solubility product expressions for AgI and PbI鈧

Write the balanced equations for the dissociation of AgI and PbI鈧 in solution and their corresponding solubility product expressions. The balanced equation for AgI dissociation: \(AgI_{(s)} \rightleftharpoons Ag_{(aq)}^{+} + I_{(aq)}^{-}\) The solubility product expression for AgI: \( K_{sp, AgI} = [Ag^{+}][I^{-}]\) The balanced equation for PbI鈧 dissociation: \(PbI_{2(s)} \rightleftharpoons Pb_{(aq)}^{2+} + 2I_{(aq)}^{-}\) The solubility product expression for PbI鈧: \( K_{sp, PbI_2} = [Pb^{2+}][I^{-}]^2\)
02

Calculate the reaction quotient (Q) for AgI and PbI鈧

Calculate the reaction quotient (Q) for AgI and PbI鈧 using the given concentrations of Ag鈦 and Pb虏鈦. We'll assume iodide-to-be-added has a molar concentration of "\([I^{-}]\)" for simplicity. For AgI: \(Q_{AgI} = [Ag^{+}][I^{-}] = (2.0 \times 10^{-4})([I^{-}])\) For PbI鈧: \(Q_{PbI_2} = [Pb^{2+}][I^{-}]^2 = (1.5 \times 10^{-3})([I^{-}])^2\)
03

Compare Q to Ksp for both precipitates

To determine which compound precipitates first, compare the reaction quotient (Q) of each compound to the relative solubility product (Ksp). The compound that precipitates first is the one where the value of Q_for_added_iodide is greater than or equal to Ksp For AgI: \(Q_{AgI} = (2.0 \times 10^{-4})([I^{-}]) \geq 8.3 \times 10^{-17}\), \([I^{-}] \geq \frac{8.3 \times 10^{-17}}{2.0 \times 10^{-4}} = 4.15 \times 10^{-13}\) For PbI鈧: \(Q_{PbI_2} = (1.5 \times 10^{-3})([I^{-}])^2 \geq 7.9 \times 10^{-9}\), \([I^{-}] \geq \sqrt{\frac{7.9 \times 10^{-9}}{1.5 \times 10^{-3}}} = 7.3 \times 10^{-4}\)
04

Determine which precipitate forms first and the iodide concentration required

Based on the values calculated in Step 3, we can now determine which precipitate forms first and the minimum iodide concentration required to begin precipitation. Since AgI has a smaller iodide concentration required for precipitation, AgI will precipitate first. Minimum concentration of \(I^-\) required to begin precipitation: \([I^{-}] = 4.15 \times 10^{-13} M\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

AgI Precipitation
AgI, or silver iodide, precipitation involves the formation of a solid compound from ions in a solution. This occurs when the product of the concentrations of the ions exceeds the solubility product constant, known as the solubility product, or \(K_{sp}\). For AgI, the dissociation is described by:
  • \(AgI_{(s)} \rightleftharpoons Ag_{(aq)}^{+} + I_{(aq)}^{-}\)
The solubility product expression is \(K_{sp, AgI} = [Ag^{+}][I^{-}]\).When sodium iodide (NaI) is added to a solution containing silver ions \((Ag^+)\), silver iodide will begin to precipitate when \([Ag^+][I^-] \geq K_{sp}\). In this scenario, with \(K_{sp} = 8.3 \times 10^{-17}\), precipitation of AgI starts at a very low iodide concentration, meaning it generally occurs first before other possible precipitates like lead(II) iodide (PbI鈧), given the same iodide increase.
Reaction Quotient (Q)
The reaction quotient, \(Q\), is similar to the equilibrium constant, \(K\), but it applies to non-equilibrium conditions. It helps us understand whether a reaction will proceed forward or backward to reach equilibrium. For a solution, \(Q\) predicts if precipitation will occur:
  • \(Q < K_{sp}\): The solution is unsaturated, and no precipitation occurs.
  • \(Q = K_{sp}\): The solution is saturated; it is at the point of precipitation equilibrium.
  • \(Q > K_{sp}\): The solution is supersaturated, and precipitation occurs to reduce ion concentration.
In the exercise, the \(Q\) values for AgI and PbI鈧 are calculated using initial ion concentrations, indicating which compound will precipitate first as iodide ions increase. AgI precipitates before PbI鈧 because its required \(Q\), based on \(K_{sp}\), is exceeded at a lower \([I^-]\). This shows that the order of \(Q\) meeting or exceeding \(K_{sp}\) dictates the sequence of precipitation.
Iodide Concentration
The iodide concentration \([I^-]\) plays a crucial role in determining when a precipitate will form in a solution, specifically for the salts AgI and PbI鈧. The goal is to find the concentration threshold at which the product of the ion concentrations meets or exceeds the solubility product constant, \(K_{sp}\). For silver iodide (AgI), the minimum concentration of \([I^-]\) required to begin precipitation is calculated as:
  • \([I^-] \geq \frac{K_{sp}}{[Ag^+]} = \frac{8.3 \times 10^{-17}}{2.0 \times 10^{-4}} = 4.15 \times 10^{-13} \text{ M}\)
This very low concentration of iodide means AgI will precipitate first before PbI鈧 in a scenario where iodide is gradually added. For PbI鈧, a higher iodide concentration of \(7.3 \times 10^{-4} \text{ M}\) is necessary for precipitation. Thus, knowledge of \([I^-]\) enables careful control of the precipitation process in solutions containing multiple ion species.

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Most popular questions from this chapter

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) is adjusted to \(8.0 ?\) (b) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when \(100 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(5.0 \times 10^{-2} \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

(a) Why is the concentration of undissolved solid not explicitly included in the expression for the solubilityproduct constant? (b) Write the expression for the solubility-product constant for each of the following strong electrolytes: \(\mathrm{AgI}, \mathrm{SrSO}_{4}, \mathrm{Fe}(\mathrm{OH})_{2}\), and \(\mathrm{Hg}_{2} \mathrm{Br}_{2}\).

To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\) \(\left(K_{s p}=4 \times 10^{-10}\right)\) in \(1.0 \mathrm{~L}\) of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) because the solution will be quite basic.)

Suppose that a 10-mL sample of a solution is to be tested for \(\mathrm{Cl}^{-}\) ion by addition of 1 drop \((0.2 \mathrm{~mL})\) of \(0.10 \mathrm{M}\) \(\mathrm{AgNO}_{3}\). What is the minimum number of grams of \(\mathrm{Cl}^{-}\) that must be present for \(\mathrm{AgCl}(\mathrm{s})\) to form?

A buffer is prepared by adding \(20.0 \mathrm{~g}\) of acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) and \(20.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to enough water to form \(2.00 \mathrm{~L}\) of solution. (a) Determine the \(\mathrm{pH}\) of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

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