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Chapter 16: Problem 38

Calculate the molar solubility of $\mathrm{Co}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2.5 \times 10^{-43}.$

Short Answer

Expert verified
The molar solubility of \(Co(OH)_{3}\) is approximately \(3.0 \times 10^{-11}\) mol/L.

Step by step solution

01

Write the balanced dissociation equation for Co(OH)3

First, we need to write the balanced equation for the dissociation of Co(OH)3 in water. When Co(OH)3 dissolves, it dissociates into Co3+ and OH- ions: \[Co(OH)_{3}(s) \rightleftharpoons Co^{3+}(aq) + 3OH^-(aq)\]
02

Set up an ICE table

An ICE (Initial, Change, Equilibrium) table helps us keep track of the initial concentrations, changes due to dissociation, and equilibrium concentrations of the species involved in the dissociation. In this case, for the dissociation of Co(OH)3: \[\begin{array}{c|ccc} & Co(OH)_{3}& Co^{3+} & 3OH^{-} \\ \hline I & \text{solid} & 0 & 0 \\ C & & +s & +3s \\ E & & s & 3s \\ \end{array}\] Here, 's' represents the molar solubility of Co(OH)3, which we are trying to find.
03

Write the expression for Ksp

To find the solubility product constant, Ksp, we will write the expression based on the equilibrium concentrations of the ions in the dissociation equation: \[K_{sp} = [Co^{3+}] [OH^{-}]^3\] Plug the equilibrium concentrations from the ICE table into the equation: \[K_{sp} = (s)(3s)^3\]
04

Solve for molar solubility 's'

Now we have the equation to find the molar solubility, using the given Ksp value, 2.5 x 10^-43: \[K_{sp} = 2.5 \times 10^{-43} = (s)(3s)^3\] First, simplify the equation: \[2.5 \times 10^{-43} = 27s^4\] Now, solve for 's': \[s^4 = \frac{2.5 \times 10^{-43}}{27}\] \[s = \sqrt[4]{\frac{2.5 \times 10^{-43}}{27}}\] \[ s \approx 3.0 \times 10^{-11}\] So, the molar solubility of Co(OH)3 is approximately 3.0 x 10^-11 mol/L.

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Most popular questions from this chapter

The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is 1 $\mathrm{mg} \mathrm{F}^{-}$ per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level? $\left(K_{\mathrm{sp}} \text { for } \mathrm{CaF}_{2}=4.0 \times 10^{-11}\right)$

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of $\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}$ b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{ED} \mathrm{TA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+} \mathrm{is}\) $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(\mathrm{M}\) Na_thion. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

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 $K_{\mathrm{sp}} \mathrm{for} \mathrm{MnS}=2.3 \times 10^{-13} .$

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.

The \(K_{\mathrm{sp}}\) for \(Q,\) a slightly soluble ionic compound composed of \(\mathrm{M}_{2}^{2+}\) and \(\mathrm{X}^{-}\) ions, is \(4.5 \times 10^{-29} .\) The electron configuration of \(\mathrm{M}^{+}\) is $[\mathrm{Xe}] 6 s^{1} 4 f^{4} 5 d^{10} .\( The \)\mathrm{X}^{-}$ anion has 54 electrons. What is the molar solubility of \(Q\) in a solution of \(\mathrm{NaX}\) prepared by dissolving \(1.98 \mathrm{g}\) \(\mathrm{NaX}\) in \(150 .\) mL solution?
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