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Chapter 16: Problem 111

What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10 M\) $\mathrm{H}_{2} \mathrm{S}\( and maintained at \)\mathrm{pH} 3.0\( with \)\mathrm{HCl} ?$

Short Answer

Expert verified
The maximum concentration of \(\mathrm{Ni}^{2+}\) ion in the saturated solution is \(5.5 \times 10^{-12} M\).

Step by step solution

01

Write the balanced equation for the reaction involving the given species.

\(\mathrm{Ni}^{2+} + \mathrm{H}_{2}\mathrm{S} \rightleftharpoons \mathrm{NiS}\downarrow + 2\mathrm{H}^{+}\) #Step 2: Write the equilibrium constant expression#
02

Write the equilibrium constant expression for the reaction.

For the reaction, the equilibrium constant is given by: \(K = \frac{[\mathrm{NiS}][\mathrm{H}^{+}]^{2}}{[\mathrm{Ni}^{2+}][\mathrm{H}_{2}\mathrm{S}]}\) #Step 3: Find the concentration of \(\mathrm{H}^{+}\) from the given pH.#
03

Calculate the concentration of \(\mathrm{H}^{+}\) ions based on the given pH.

The concentration of \(\mathrm{H}^{+}\) is given by: \([\mathrm{H}^{+}]= 10^{-\mathrm{pH}}\) Plug in the given pH value: \([\mathrm{H}^{+}]= 10^{-3}\) #Step 4: Determine the solubility product constant, \(K_{sp}\), of \(\mathrm{NiS}\).#
04

Find the solubility product constant for \(\mathrm{NiS}\) from a reference.

From the reference, we find that the solubility product constant for \(\mathrm{NiS}\) is \(K_{sp} = 5.5 \times 10^{-14}\). #Step 5: Calculate the concentration of \(\mathrm{Ni}^{2+}\) using the equilibrium constant expression and given data.#
05

Use the values of \(K_{sp}\), \([\mathrm{H}_{2}\mathrm{S}]\), and \([\mathrm{H}^{+}]\) to find the concentration of \(\mathrm{Ni}^{2+}\) ions.

Since at equilibrium, \([\mathrm{NiS}] = [\mathrm{Ni}^{2+}]\), we can rewrite the equilibrium constant equation as follows: \[K_{sp} = \frac{[\mathrm{Ni}^{2+}][\mathrm{H}^{+}]^{2}}{[\mathrm{Ni}^{2+}][\mathrm{H}_{2}\mathrm{S}]}\] Plug in the values: \(5.5 \times 10^{-14} = \frac{[\mathrm{Ni}^{2+}](10^{-3})^{2}}{[\mathrm{Ni}^{2+}](0.10)}\) Solving for \([\mathrm{Ni}^{2+}]\), we get: \([\mathrm{Ni}^{2+}] = 5.5 \times 10^{-12} M\) #Step 6: Write the final answer#
06

Present the final answer for the maximum concentration of \(\mathrm{Ni}^{2+}\) ions in the saturated solution.

The maximum concentration of \(\mathrm{Ni}^{2+}\) ion in the saturated solution is \(5.5 \times 10^{-12} M\).

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Most popular questions from this chapter

A mixture contains \(1.0 \times 10^{-3} M \mathrm{Cu}^{2+}\) and $1.0 \times 10^{-3} M$ \(\mathrm{Mn}^{2+}\) and is saturated with 0.10\(M \mathrm{H}_{2} \mathrm{S} .\) Determine a pH where CuS precipitates but MnS does not precipitate. \(K_{\mathrm{sp}}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and $K_{\mathrm{sp}} \mathrm{for} \mathrm{MnS}=2.3 \times 10^{-13} .$

A solution contains \(1.0 \times 10^{-5} M \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What concentrations of \(\mathrm{A} \mathrm{g} \mathrm{NO}_{3}\) will cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\) \(\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)

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

Tooth enamel is composed of the mineral hydroxyapatite. The \(K_{\mathrm{sp}}\) of hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\( is \)6.8 \times 10^{-37}$ . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\) , forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times 10^{-60}\) . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

\(K_{\mathrm{f}}\) for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is $1.7 \times 10^{7} . K_{\mathrm{sp}}\( for \)\mathrm{AgCl}\( is \)1.6 \times 10^{-10} .$ Calculate the molar solubility of \(\mathrm{AgCl}\) in \(1.0M\) \(\mathrm{NH}_{3}.\)
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