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Chapter 16: Problem 65

The overall formation constant for \(\mathrm{HgI}_{4}{ }^{2-}\) is \(1.0 \times 10^{30} .\) That is, $$1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[1^{-}\right]^{4}}$$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{~mL}\) of a solution that was originally \(0.010 \mathrm{M} \mathrm{Hg}^{2+}\) and \(0.78 \mathrm{M} \mathrm{I}^{-}\) ? The reaction is $$\mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}{ }^{2-}(a q)$$

Short Answer

Expert verified
The final concentration of Hg²⁺ in the 500.0 mL solution at equilibrium is approximately \(9.8075 \times 10^{-3} \mathrm{M}\).

Step by step solution

01

Write the expression for the reaction quotient (Q)

For the reaction, $$\mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}{ }^{2-}(a q)$$ the reaction quotient Q can be written as, $$Q = \frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^4}$$
02

Set up an ICE table (Initial, Change, Equilibrium) for concentrations

We will use an ICE table to keep track of the concentrations of each species involved in the reaction as it progresses from the initial state to equilibrium. The initial concentrations are given: [Hg²⁺]₀ = 0.010M and [I⁻]₀ = 0.78 M`. Since HgI₄²⁻ is not initially present, [HgI₄²⁻]₀ = 0. | Species | Initial | Change | Equilibrium | |:---------:|:--------:|:--------:|:-----------:| | Hg²⁺(aq) | 0.010 | -x | 0.010-x | | I⁻(aq) | 0.78 | -4x | 0.78-4x | | HgI₄²⁻(aq) | 0 | +x | x | The changes in the concentrations are multiplied by stoichiometric coefficients (i.e., -x for Hg²⁺, -4x for I⁻, and +x for HgI₄²⁻).
03

Substitute equilibrium concentrations in Q expression

Now, we'll plug the equilibrium concentrations from the ICE table into the reaction quotient expression: $$Q = \frac{x}{(0.010-x)(0.78-4x)^4}$$
04

Set Q equal to the equilibrium constant (Kf) and solve for x

At equilibrium, Q is equal to the formation constant Kf, so we can set up the following equation: $$Kf = \frac{x}{(0.010-x)(0.78-4x)^4}=1.0 \times 10^{30}$$ This equation is difficult to solve directly since it involves high powers of x. However, given the magnitude of Kf, we can assume that the change in concentration of Hg²⁺ (x) is negligible compared to the initial concentration (0.010M). This means we can approximate 0.010-x as 0.010 and 0.78-4x as 0.78. Now we can rewrite the equation: $$1.0 \times 10^{30}=\frac{x}{(0.010)(0.78)^4}$$
05

Solve for x and find the concentration of Hg²⁺ at equilibrium

We now solve this simplified equation for x: $$x = 1.0 \times 10^{30} \times (0.010)(0.78)^4$$ By calculating the value of x, we get: $$x = 1.925 \times 10^{-5}$$ The equilibrium concentration of Hg²⁺ is [Hg²⁺] = 0.010 - x, which is approximately: $$[Hg²⁺] \approx 0.010 - 1.925 \times 10^{-5} = 9.8075 \times 10^{-3} \mathrm{M}$$ So, the final concentration of Hg²⁺ in the 500.0 mL solution at equilibrium is approximately 9.8075 × 10⁻³ M.

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Most popular questions from this chapter

a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\), calculate the value for the equilibrium constant for the following reaction: \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 \mathrm{M}\) \(\mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 \mathrm{M} .\)

For which salt in each of the following groups will the solubility depend on pH? a. \(\mathrm{AgF}, \mathrm{AgCl}, \mathrm{AgBr}\) c. \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Sr}\left(\mathrm{NO}_{2}\right)_{2}\) b. \(\mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}\) d. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}\)

The stepwise formation constants for a complex ion are all generally values much greater than 1 . What is the significance of this?

The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times 10^{-6}\). Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}\) : $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 \mathrm{M} \mathrm{NaCl}\).

Calculate the final concentrations of \(\mathrm{K}^{+}(a q), \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)\), \(\mathrm{Ba}^{2+}(a q)\), and \(\mathrm{Br}^{-}(a q)\) in a solution prepared by adding \(0.100 \mathrm{~L}\) of \(0.200 \mathrm{M} \mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to \(0.150 \mathrm{~L}\) of \(0.250 \mathrm{M} \mathrm{BaBr}_{2}\). (For \(\mathrm{BaC}_{2} \mathrm{O}_{4}\) \(\left.K_{\mathrm{sp}}=2.3 \times 10^{-\mathrm{s}} .\right)\)
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