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Chapter 18: Problem 30

What is the minimum \(\mathrm{pH}\) at which \(\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})\) will precipitate from a solution that is \(0.086 \mathrm{M}\) in \(\mathrm{Cr}^{3+}(\mathrm{aq}) ?\)

Short Answer

Expert verified
The minimum pH at which Cr(OH)3 will precipitate from the solution is 5.18

Step by step solution

01

Write the equilibrium reaction and the expression for \(K_{sp}\)

The equilibrium reaction for \(\mathrm{Cr(OH)_3(s)}\) is given by \(\mathrm{Cr(OH)_3(s)} \leftrightharpoons \mathrm{Cr^{3+}(aq)} + 3\mathrm{OH^{-}(aq)}\)Our \(K_{sp}\) equation is: \(K_{sp} = [\mathrm{Cr^{3+}}][\mathrm{OH^-}]^3\)Assuming \(\mathrm{Cr(OH)_3(s)}\) is slightly soluble, then \([\mathrm{Cr^{3+}}]\) is equal to the original concentration of \(\mathrm{Cr^{3+}}\) in the question which is 0.086 M.
02

Determine the \(K_{sp}\) for \(\mathrm{Cr(OH)_3(s)}\)

The solubility product constant, \(K_{sp}\), describes the equilibrium between a solid and its constituent ions in a solution. The value of \(K_{sp}\) for \(\mathrm{Cr(OH)_3(s)}\) can be looked up from a textbook or online and is usually given as \(6.3 × 10^{−31}\). We then solve for \([\mathrm{OH^-}]\) from our \(K_{sp}\) equation\(6.3 × 10^{−31} = 0.086 × [\mathrm{OH^-}]^3\)
03

Solve for \([\mathrm{OH^-}]\) and find the \(\mathrm{pOH}\)

We can rearrange and solve for \([\mathrm{OH^-}]\): \([\mathrm{OH^-}] = \sqrt[3]{\frac{6.3 × 10^{-31}}{0.086}} = 1.5 × 10^{-9} M\)From the definition of pOH, \(\mathrm{pOH} = - \log([\mathrm{OH^-}])\), we substitute and calculate to get a pOH value of 8.82.
04

Calculate the pH

Finally, knowing that \(pH + pOH = 14\) at 25°C, we can calculate the pH:\(pH = 14 - pOH = 14 - 8.82 = 5.18\)

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Most popular questions from this chapter

In your own words, define the following terms or symbols: (a) \(K_{\mathrm{sp}} ;\) (b) \(K_{f} ;\) (c) \(Q_{\mathrm{sp}} ;\) (d) complex ion.

Adding \(1.85 \mathrm{g} \mathrm{Na}_{2} \mathrm{SO}_{4}\) to \(500.0 \mathrm{mL}\) of saturated aqueous \(\mathrm{BaSO}_{4}:\) (a) reduces \(\left[\mathrm{Ba}^{2+}\right] ;\) (b) reduces \(\left[\mathrm{SO}_{4}^{2-}\right]\); (c) increases the solubility of \(\mathrm{BaSO}_{4} ;\) (d) has no effect.

For the equilibrium \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})\) \(K_{\mathrm{sp}}=1.3 \times 10^{-33}\) (a) What is the minimum \(p H\) at which \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) will precipitate from a solution that is \(0.075 \mathrm{M}\) in \(\mathrm{Al}^{3+} ?\) (b) A solution has \(\left[\mathrm{Al}^{3+}\right]=0.075 \mathrm{M}\) and \(\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=1.00 \mathrm{M} .\) What is the maximum quantity of \(\mathrm{NaCH}_{3} \mathrm{COO}\) that can be added to \(250.0 \mathrm{mL}\) of this solution before precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) begins?

A \(250 \mathrm{mL}\) sample of saturated \(\mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{aq})\) requires \(4.8 \mathrm{mL}\) of \(0.00134 \mathrm{M} \mathrm{KMnO}_{4}(\mathrm{aq})\) for its titration in an acidic solution. What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) obtained with these data? In the titration reaction, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is oxidized to \(\mathrm{CO}_{2}\) and \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\).

Suppose you did a group 1 qualitative cation analysis and treated the chloride precipitate with \(\mathrm{NH}_{3}(\mathrm{aq})\) without first treating it with hot water. What might you observe, and what valid conclusions could you reach about cations present, cations absent, and cations in doubt?
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