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Chapter 14: Problem 95

At \(25^{\circ} \mathrm{C}\), the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in water is \(2.86 \times 10^{-9} \mathrm{M}\). What are the equilibrium concentrations of the cation and the anion in a saturated solution?

Short Answer

Expert verified
In a saturated solution of Al(OH)3 at 25°C with solubility \(2.86 \times 10^{-9} \mathrm{M}\), the equilibrium concentrations of Al3+ and OH- are \(2.86 \times 10^{-9} \mathrm{M}\) and \(8.58 \times 10^{-9} \mathrm{M}\), respectively.

Step by step solution

01

Write the balanced chemical equation for Al(OH)3's dissolution

Al(OH)3 (s) ↔ Al3+ (aq) + 3 OH- (aq)
02

Set up the expression for Ksp

For a balanced chemical equation 'aA(s) ↔ bB(aq) + cC(aq)', the solubility product constant (Ksp) can be defined as: Ksp = [B]^b × [C]^c Using the balanced chemical equation for the dissolution of Al(OH)3, Ksp = [Al3+] × [OH-]^3
03

Calculate the equilibrium concentrations of cation and anion

Since the solubility of Al(OH)3 is given as \(2.86 \times 10^{-9}\) M, we have the same solubility for Al3+ as it is in a 1:1 stoichiometry with Al(OH)3. Therefore, [Al3+] = \(2.86 \times 10^{-9}\) M. Now, the balanced chemical equation shows that there is a 1:3 stoichiometry between Al3+ and OH-. Hence, for each Al3+ ion, there will be three OH- ions in the solution. Thus, to find [OH-], we multiply the given molar concentration of Al3+ by 3: [OH-] = 3 × [Al3+] = 3 × \(2.86 \times 10^{-9}\) M = \(8.58 \times 10^{-9}\) M Therefore, the equilibrium concentrations of Al3+ and OH- are \(2.86 \times 10^{-9}\) M and \(8.58 \times 10^{-9}\) M, respectively.

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Most popular questions from this chapter

Sometimes reactions are written with two arrows pointing in opposite directions instead of a single arrow going from reactants to products. What do the two arrows mean?

How does decreasing the temperature affect the value of \(K_{e q}\) for an exothermic reaction? (a) Increases \(K_{\text {eq }}\) (b) Decreases \(K_{\text {eq }}\) (c) Does not change \(K_{\text {eq }}\)

\(K_{\text {eq }}=3.9 \times 10^{-11}\) for the dissolution of calcium fluoride in water: \(\mathrm{CaF}_{2}(s) \rightleftarrows \mathrm{Ca}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)\) (a) What is another name for \(K_{\mathrm{eq}}\) for this reaction? (b) If the equilibrium calcium ion concentration in a saturated aqueous solution of calcium fluoride is \(3.3 \times 10^{-4} \mathrm{M}\), what is the equilibrium fluoride ion concentration? (c) Which is larger, the rate constant for the forward reaction or the rate constant for the reverse reaction? (d) Which is larger, \(E_{a}\) for the forward reaction or \(E_{\mathrm{a}}\) for the reverse reaction? (e) Which is larger, the rate of the forward reaction or the rate of the reverse reaction? (f) For lithium carbonate, \(K_{\mathrm{sp}}=0.0011\). Write the balanced chemical equation and the equilibrium expression for the dissolution of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) in water. (g) Which is more soluble in water, calcium fluoride or lithium carbonate?

A chemist runs a reaction that is known to proceed very rapidly and keeps isolating product that is contaminated with starting material, even though he is following the stoichiometry of the reaction and giving the reaction enough time to run. How might this be explained?

Suppose a reaction is at equilibrium and you then disturb the equilibrium by adding reactants. What happens to the value of \(K_{\text {eq }}\) ? Explain your answer.
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