Problem 69 - Chemistry: The Central Science in SI Units, Global Edition

Question

(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 \(8.0 ?(\mathbf{b})\) 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?

Step-by-Step Solution

Answer

In summary, \( Ca(OH)_2 \) will not precipitate from a 0.050 M solution of \( CaCl_2 \) when the pH is adjusted to 8.0 because the reaction quotient (Q) is much less than the solubility product constant (Ksp). Similarly, \( Ag_2SO_4 \) will not precipitate when 100 mL of 0.050 M \( AgNO_3 \) is mixed with 10 mL of \( 5.0 \times 10^{-2} M \, Na_2SO_4 \) solution, as Q is less than Ksp in this case as well.

Step 1: Calculate the concentration of OH- ions

The pH of the solution is given as 8.0, so we can find the concentration of OH- ions using the following relation: pOH = 14 - pH Therefore, pOH = 14 - 8 = 6 By definition, pOH = -log10[OH-]. So, we can calculate the concentration of OH- ions as: [OH-] = 10^(-pOH) = 10^(-6) = 1.0 * 10^(-6) M

Step 2: Calculate the reaction quotient, Q

Now, we'll calculate the reaction quotient, Q, for the reaction: Ca(OH)2(s) <=> Ca2+(aq) + 2OH-(aq) In this case, Q = [Ca2+][OH-]^2, where [Ca2+] is the concentration of Ca2+ ions and [OH-] is the concentration of OH- ions. Given that [Ca2+] = 0.050 M (from the CaCl2 solution) and [OH-] = 1 * 10^(-6) M (from the pH), we can calculate Q as: Q = (0.050)(1.0 * 10^(-6))^2 = 5.0 * 10^(-13)

Step 3: Compare Q to the solubility product constant, Ksp

The solubility product constant, Ksp, for Ca(OH)2 is 6.5 * 10^(-6). Let's compare Q to Ksp to determine if Ca(OH)2 will precipitate from the solution. Since Q (5.0 * 10^(-13)) is much less than Ksp (6.5 * 10^(-6)), no precipitate will form. Therefore, Ca(OH)2 will not precipitate from the solution when the pH is adjusted to 8.0. (b) Will Ag2SO4 precipitate when 100 mL of 0.050 M AgNO3 is mixed with 10 mL of 5.0 * 10^(-2) M Na2SO4 solution?

Step 4: Calculate the concentrations of Ag+ and SO4 2- ions

We can find the concentrations of Ag+ and SO4 2- ions after mixing the solutions as follows: [Ag+] = (0.050 M * 100 mL) / (100 mL + 10 mL) = 0.0455 M [SO4 2-] = (5.0 * 10^(-2) M * 10 mL) / (100 mL + 10 mL) = 4.55 * 10^(-3) M

Step 5: Calculate the reaction quotient, Q

Now, we'll calculate the reaction quotient, Q, for the reaction: Ag2SO4(s) <=> 2Ag+(aq) + SO4 2-(aq) In this case, Q = [Ag+]^2[SO4 2-], so we can calculate Q as: Q = (0.0455)^2(4.55 * 10^(-3)) = 9.38 * 10^(-6)

Step 6: Compare Q to the solubility product constant, Ksp

The solubility product constant, Ksp, for Ag2SO4 is 1.2 * 10^(-5). Let's compare Q to Ksp to determine if Ag2SO4 will precipitate from the solutions. Since Q (9.38 * 10^(-6)) is less than Ksp (1.2 * 10^(-5)), no precipitate will form. Therefore, Ag2SO4 will not precipitate when the given solutions are mixed.

Key Concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, often denoted as \( K_{sp} \), is a critical value in predicting whether a compound will dissolve or precipitate in a solution. It represents the equilibrium constant for a solid substance dissolving in an aqueous solution. The \( K_{sp} \) value varies with temperature and is unique for each sparingly soluble compound. Understanding \( K_{sp} \) is essential for solving chemistry problems involving solution equilibria.

For any dissolution reaction, such as \( AB(s) \rightleftharpoons A^{+}(aq) + B^{-}(aq) \), the solubility product is expressed as: \( K_{sp} = [A^{+}][B^{-}] \). A higher \( K_{sp} \) indicates greater solubility, whereas a lower \( K_{sp} \) suggests lower solubility.

In practical terms:

• If \( Q = K_{sp} \), the solution is at equilibrium, meaning the maximum amount of solute has dissolved.
• If \( Q < K_{sp} \), more solute can dissolve in the solution.
• If \( Q > K_{sp} \), the excess solute will precipitate out of the solution.

Understanding \( K_{sp} \) helps predict precipitation reactions, which are fundamental in lab procedures and industrial processes.

Reaction Quotient (Q)

The Reaction Quotient, \( Q \), is a snapshot of a system's status at a specific moment in time, concerning reactants and products. It is calculated in the same way as the equilibrium constant, \( K \), but does not require the system to be at equilibrium.

For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the expression for \( Q \) is given by: \[ Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
This formula uses molarities of the substances in the reaction at any particular point in time.

If we consider the solutions given in the prompt, calculating \( Q \) for the proposed conditions tells us how close the system is to equilibrium:

• When \( Q < K_{sp} \), the solution will dissolve more solute until \( Q \) equals \( K_{sp} \).
• If \( Q = K_{sp} \), the system is at equilibrium.
• When \( Q > K_{sp} \), there is too much solute in solution, and it will precipitate to restore balance.

The comparison between \( Q \) and \( K_{sp} \) helps in deciding whether a precipitation reaction is expected to occur.

Precipitation Reactions

Precipitation reactions occur when two solutions combine to form an insoluble solid, known as a precipitate. These reactions are fundamental in various analytical and separation techniques in chemistry.

There are several factors that influence whether a precipitation reaction will take place:

• Solubility of the species: The solubility rules and the \( K_{sp} \) value dictate which combinations of ions form precipitates.
• Concentration of ions: When the concentration of resulting ions fits the condition \( Q > K_{sp} \), a precipitate forms.
• Temperature: Changes in temperature can affect solubility and influence precipitation.

To predict precipitation, we evaluate and compare \( Q \) with \( K_{sp} \). If mixing the solutions results in \( Q \) greater than \( K_{sp} \), a precipitate will form as seen in the examples, such as the Aguilar sulfate solution in the steps.

Knowing how and when precipitation occurs is crucial not just in academic problems but also in real-world applications, like waste treatment and resource extraction processes. Understanding these concepts enables chemists to manipulate experimental conditions to produce desired results.