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Chapter 16: Problem 78

The copper(I) ion forms a chloride salt that has $K_{\mathrm{sp}}= 1.2 \times 10^{-6} .\( Copper (I) also forms a complex ion with \)\mathrm{Cl}^{-} :$ $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \qquad K=8.7 \times 10^{4}$$ 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\(M\) \(\mathrm{NaCl}\).

Short Answer

Expert verified
The solubility of copper(I) chloride in pure water is about \(1.1\times10^{-3}\,\mathrm{M}\) and in a 0.10\(M\) \(\mathrm{NaCl}\) solution, it is approximately \(8.7\times10^{-3}\,\mathrm{M}\).

Step by step solution

01

Write the chemical equation and the solubility product expression for the dissolution of copper(I) chloride

The dissolution of copper(I) chloride in water can be represented as follows: \[\mathrm{CuCl}(s) \rightleftharpoons \mathrm{Cu}^{+}(a q) + \mathrm{Cl}^{-}(a q)\] The solubility product expression for this reaction is given by: \[K_{\mathrm{sp}} = [\mathrm{Cu}^{+}][\mathrm{Cl}^{-}]\]
02

Find the equilibrium concentrations and set up the \(K_{\mathrm{sp}}\) equation

Let's denote the solubility of copper(I) chloride as \(s\) moles per liter. At equilibrium, the concentration of \(\mathrm{Cu}^{+}\) and \(\mathrm{Cl}^{-}\) ions will also represent \(s\). The \(K_{\mathrm{sp}}\) expression can now be expressed as: \[1.2\times10^{-6} = (s)(s)\]
03

Solve for solubility \(s\)

Now we can solve for the solubility of copper(I) chloride, \(s\). \[s^2 = 1.2\times10^{-6}\] \[s = \sqrt{1.2\times10^{-6}} \approx 1.1\times10^{-3}\,\mathrm{M}\] The solubility of copper(I) chloride in pure water is about \(1.1\times10^{-3}\,\mathrm{M}\). Part (b): Calculate the solubility of copper(I) chloride in 0.10\(M\) \(\mathrm{NaCl}\)
04

Write the chemical equation for the complex ion formation and the equilibrium expression

The formation of the copper(I) chloride complex ion with \(\mathrm{Cl}^-\) can be represented as follows: \[\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q)\] The equilibrium expression for this reaction is given by: \[K = \frac{[\mathrm{CuCl}_{2}^{-}]}{[\mathrm{Cu}^{+}][\mathrm{Cl}^{-}]^2}\]
05

Calculate the initial concentration of \(\mathrm{Cl}^-\) ions

In a 0.10\(M\) \(\mathrm{NaCl}\) solution, the initial concentration of \(\mathrm{Cl}^-\) ions will be \(0.10\,\mathrm{M}\) due to the complete dissociation of \(\mathrm{NaCl}\). Also, note that the initial concentration of \(\mathrm{Cu}^{+}\) ions and the complex ion, \(\mathrm{CuCl}_{2}^{-}\), is zero.
06

Set up an ICE table and the equilibrium expression using variables

We will set up an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations as the reaction progresses towards equilibrium. The formation of the complex ion is described by the reaction: \[\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q)\] At equilibrium, let the concentration of the complex ion \(\mathrm{CuCl}_{2}^{-}\) be \(x\,\mathrm{M}\). Then the equilibrium concentrations would be: \[[\mathrm{Cu}^{+}]=(s - x)\,\mathrm{M}\] \[[\mathrm{Cl}^{-}]=(0.10 - 2x)\,\mathrm{M}\] \[[\mathrm{CuCl}_{2}^{-}]=x\,\mathrm{M}\] Now we can write the equilibrium expression in terms of s and x: \[8.7\times10^{4} = \frac{x}{(s - x)(0.10 -2x)^2}\]
07

Solve for solubility \(s\) in 0.10\(M\) \(\mathrm{NaCl}\)

Since \(s\) from part (a) is very small compared to \(0.10\,\mathrm{M}\) (about \(1.1\times10^{-3}\,\mathrm{M}\) << \(0.10\,\mathrm{M}\)), we can assume that the term \(2x\) is negligible compared to \(0.10\,\mathrm{M}\). Therefore, the equilibrium expression becomes: \[8.7\times10^{4} \approx \frac{x}{(s - x)(0.10)^2}\] Now, we can calculate \(\frac{s}{0.10^2}\) knowing that \(\frac{[Cl^-]{[Cu^{+}]}}{[CuCl_{2}^{-}]}\) : \[\frac{s}{0.10^2} = \frac{[Cl^-]{[Cu^{+}]}}{[CuCl_{2}^{-}]} =8.7\times10^{4} \] \(s = 0.10^2 \times 8.7\times10^{4}\) \(s \approx 8.7\times10^{-3}\,\mathrm{M}\) The solubility of copper(I) chloride in 0.10\(M\) \(\mathrm{NaCl}\) is approximately \(8.7\times10^{-3}\,\mathrm{M}\).

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Most popular questions from this chapter

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-} .\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are $8.5 \times 10^{-40} \mathrm{M}\( and \)1.5 \times 10^{-3} M,\( respectively, in a \)0.11-M$ KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$

A friend tells you: "The constant \(K_{\mathrm{sp}}\) of a salt is called the solubility product constant and is calculated from the concentrations of ions in the solution. Thus, if salt A dissolves to a greater extent than salt \(\mathrm{B}\) , salt \(\mathrm{A}\) must have a higher \(K_{\mathrm{sp}}\) than salt \(\mathrm{B}\) ." Do you agree with your friend? Explain.

Calculate the molar solubility of $\mathrm{Al}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2 \times 10^{-32}.$

The copper(l) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: $$\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ a. Calculate the solubility of $\mathrm{CuBr}(s)\left(K_{\mathrm{sp}}=1.0 \times 10^{-5}\right)\( in \)1.0 \mathrm{L}\( of \)1.0 \mathrm{M} \mathrm{NaCN}$ . b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.

The solubility of copper(II) hydroxide in water can be increased by adding either the base \(\mathrm{NH}_{3}\) or the acid \(\mathrm{HNO}_{3}\) . Explain. Would added \(\mathrm{NH}_{3}\) or \(\mathrm{HNO}_{3}\) have the same effect on the solubility of silver acetate or silver chloride? Explain.
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