/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Calculate the molar solubility o... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 32

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)鈧� is approximately \( 1.25 \times 10^{-11}\: \text{mol/L} \).

Step by step solution

01

1. Write the balanced chemical equation of Co(OH)鈧� dissolving in water

First, we need to know how Co(OH)鈧� behaves in the water when it dissolves. The balanced reaction for the dissolution of cobalt(III) hydroxide is: Co(OH)鈧� (s) 鈬� Co鲁鈦� (aq) + 3OH鈦� (aq)
02

2. Express the solubility in terms of x

Now, let's assume the molar solubility of Co(OH)鈧� is x mol/L. Then at equilibrium: - [Co鲁鈦篯 = x mol/L - [OH鈦籡 = 3x mol/L
03

3. Write the expression for Ksp

Now, let's write the expression for Ksp for this reaction using the molar concentrations at equilibrium: Ksp = [Co鲁鈦篯 [OH鈦籡鲁
04

4. Substitute the given Ksp value and molar concentrations in the Ksp expression

Since the question provides the Ksp value of 2.5 脳 10鈦烩伌鲁, now we can substitute the values in our expression: \( 2.5 \times 10^{-43} \) = (x) (3x)鲁
05

5. Solve for x

To find the molar solubility x, we need to simplify and solve the equation for x: \( 2.5 \times 10^{-43} \) = 27x鈦� Divide both sides by 27: \( x^{4} = \frac{2.5 \times 10^{-43}}{27} \) Now, take the fourth root of both sides: \( x = \left(\frac{2.5 \times 10^{-43}}{27}\right)^{\frac{1}{4}} \)
06

6. Calculate the molar solubility

Finally, we can plug in the given values into our calculator to find the molar solubility: \( x = \left(\frac{2.5 \times 10^{-43}}{27}\right)^{\frac{1}{4}} \approx 1.25 \times 10^{-11}\: \text{mol/L} \) So, the molar solubility of Co(OH)鈧� is approximately \( 1.25 \times 10^{-11}\: \text{mol/L} \).

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Most popular questions from this chapter

A friend tells you: "The constant \(K_{\text {sp }}\) 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.

Will a precipitate form when \(100.0 \mathrm{~mL}\) of \(4.0 \times 10^{-4} M\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(100.0 \mathrm{~mL}\) of \(2.0 \times 10^{-4} \mathrm{M} \mathrm{NaOH} ?\)

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, \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\), 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?

A solution is formed by mixing \(50.0 \mathrm{~mL}\) of \(10.0 \mathrm{M} \mathrm{NaX}\) with \(50.0 \mathrm{~mL}\) of \(2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3} .\) Assume that \(\mathrm{Cu}(\mathrm{I})\) forms com- plex ions with \(\mathrm{X}^{-}\) as follows: $$\begin{aligned}\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) & K_{1}=1.0 \times 10^{2} \\ \mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) & K_{2}=1.0 \times 10^{4} \\ \mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) & K_{3}=1.0 \times 10^{3} \end{aligned}$$ with an overall reaction $$\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K=1.0 \times 10^{9}$$ Calculate the following concentrations at equilibrium. a. \(\mathrm{CuX}_{3}^{2-}\) b. \(\mathrm{CuX}_{2}\) c. \(\mathrm{Cu}^{+}\)

Solutions of sodium thiosulfate are used to dissolve unexposed \(\mathrm{AgBr}\left(K_{\mathrm{sp}}=5.0 \times 10^{-13}\right)\) in the developing process for blackand-white film. What mass of AgBr can dissolve in \(1.00 \mathrm{~L}\) of \(0.500 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} ? \mathrm{Ag}^{+}\) reacts with \(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\) to form a complex ion:
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