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Chapter 15: Problem 67

The overall formation constant for \(\mathrm{HgI}_{4}^{2-}\) is \(1.0 \times 10^{30}\) That is, $$ 1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{4}} $$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{mL}\) of a solution that was originally \(0.010\) \(M\) \(\mathrm{Hg}^{2+}\) and \(0.78\) \(M\) \(\mathrm{I}^{-} ?\) 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 concentration of Hg^2+ in 500.0 mL of the solution is approximately \(7.07 \times 10^{-3} M\).

Step by step solution

01

Initial concentration table

Set up a preliminary Ice Table: ``` [Hg^2+] [I^-] [HgI4^2-] Initial 0.010 M 0.78 M 0 ```
02

Change in concentration

Since the reaction is Hg^2+(aq) + 4I^-(aq) <=> HgI4^2-(aq), let the change in concentration of Hg^2+ be x, and I^- be 4x. Thus, the concentration change of HgI4^2- will also be x. ``` [Hg^2+] [I^-] [HgI4^2-] Initial 0.010 M 0.78 M 0 Change -x -4x +x ```
03

Equilibrium concentration

Equilibrium concentrations are as follows: ``` [Hg^2+] [I^-] [HgI4^2-] Equilibrium 0.010-x 0.78-4x x ```
04

Formation constant equation

Write the equation for the formation constant of HgI4^2-, using the given number and equilibrium concentrations: \(1.0 \times 10^{30} = \frac{[HgI4^{2-}]}{[Hg^{2+}][I^-]^4}\) Substitute equilibrium concentrations from the table: \(1.0 \times 10^{30} = \frac{x}{(0.010-x)(0.78-4x)^4}\)
05

Simplify and solve

Make the simplifying assumption that x will be much smaller than 0.010 and 0.78, so the expression becomes: \(1.0 \times 10^{30} = \frac{x}{(0.010)(0.78)^4}\) Solve for x: \(x = (0.010)(0.78)^4 (1.0 \times 10^{30})\) \(x \approx 2.93 \times 10^{-2}\)
06

Calculate the concentration of Hg^2+

The concentration of Hg^2+ at equilibrium is (0.010-x). Substituting the value of x: [ Hg^2+ ] = (0.010 - 2.93 \times 10^{-2}) \approx 7.07 \times 10^{-3} M So, the concentration of Hg^2+ in 500.0 mL of the solution is approximately \(7.07 \times 10^{-3} M\).

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Most popular questions from this chapter

A solution contains \(1.0 \times 10^{-5} M \mathrm{Ag}^{+}\) and \(2.0 \times 10^{-6} M \mathrm{CN}^{-}\) Will AgCN( \(s\) ) precipitate? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{AgCN}(s) \text { is } 2.2 \times 10^{-12} .\right)\)

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{CaF}_{2}(s), K_{\mathrm{sp}}=4.0 \times 10^{-11},\) or \(\mathrm{BaF}_{2}(s), K_{\mathrm{sp}}=2.4 \times 10^{-5}\) b. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s), K_{\mathrm{sp}}=1.3 \times 10^{-32},\) or \(\mathrm{FePO}_{4}(s)\) \(K_{\mathrm{sp}}=1.0 \times 10^{-22}\)

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{Hg} \mathrm{I}_{4}^{2-} .\) )

The stepwise formation constants for a complex ion usually have values much greater than \(1 .\) What is the significance of this?
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