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Chapter 14: Problem 22

Sparingly soluble aluminum hydroxide dissolves in water to yield an equilibrium hydroxide ion concentration of \(8.58 \times 10^{-9} \mathrm{M}\). (a) Write the balanced equilibrium equation for aluminum hydroxide dissolving in water. (b) Write the \(K_{\mathrm{sp}}\) expression for aluminum hydroxide. (c) What is the equilibrium concentration of aluminum ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for aluminum hydroxide (show your calculation).

Short Answer

Expert verified
(a) The balanced equilibrium equation for aluminum hydroxide dissolving in water is: Al(OH)3(s) \<--> Al3+ (aq) + 3OH- (aq) (b) The Ksp expression for aluminum hydroxide is: Ksp = [Al3+][OH-]^3 (c) The equilibrium concentration of aluminum ion (Al3+) is \(2.86 \times 10^{-9} M\). (d) The value of Ksp for aluminum hydroxide is approximately \(3.00 \times 10^{-27}\).

Step by step solution

01

Write the balanced equilibrium equation

First, we need to write the balanced equilibrium equation for aluminum hydroxide dissolving in water. Aluminum hydroxide has the chemical formula Al(OH)3. When it dissolves in water, it dissociates into aluminum ions (Al3+) and hydroxide ions (OH-): Al(OH)3(s) \<--> Al3+ (aq) + 3OH- (aq)
02

Write the Ksp expression

Next, we need to write the Ksp (Solubility Product Constant) expression for aluminum hydroxide. The general formula for Ksp is the product of the concentrations of the products raised to the power of their coefficients (in the balanced equation), divided by the concentration of the reactants: Ksp = [Al3+][OH-]^3
03

Calculate the equilibrium concentration of aluminum ions

We are given that the equilibrium hydroxide ion concentration is \(8.58 \times 10^{-9} M\). Since the ratio of aluminum ions to hydroxide ions in the balanced equation is 1:3, the concentration of aluminum ions in equilibrium will be a third of the concentration of hydroxide ions. Equilibrium concentration of Al3+ = \(\frac{1}{3}\) x (equilibrium concentration of OH-) Equilibrium concentration of Al3+ = \(\frac{1}{3}\) x \(8.58 \times 10^{-9} M\) Equilibrium concentration of Al3+ = \(2.86 \times 10^{-9} M\)
04

Calculate the value of Ksp

Now that we have the equilibrium concentrations of aluminum ions and hydroxide ions, we can calculate the value of Ksp for aluminum hydroxide using the expression we found in step 2: Ksp = [Al3+][OH-]^3 Ksp = \((2.86 \times 10^{-9})(8.58 \times 10^{-9})^3\) Ksp = \(3.00 \times 10^{-27}\) The value of Ksp for aluminum hydroxide is approximately \(3.00 \times 10^{-27}\).

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Most popular questions from this chapter

Consider the gas-phase reaction \(3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{O}_{3}(g) .\) Suppose \(K_{\mathrm{eq}}\) for this reaction is \(\sim 1\) (it is not, but assume it is for this problem). Suppose you want pure ozone \(\left(\mathrm{O}_{3}\right)\) that is uncontaminated with oxygen \(\left(\mathrm{O}_{2}\right)\). (a) Why can't you simply remove the oxygen from the reaction vessel once the reaction has come to equilibrium to obtain pure ozone? (b) In fact, \(K_{\text {eq }}\) for this reaction at room temperature is \(2.5 \times 10^{-29}\). Knowing this, how important would you say Le Châtelier's principle is for this reaction when it comes to influencing the amount of ozone present at equilibrium? Explain.

In which direction does the reaction \(\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) \(\rightleftarrows \mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q)\) shift when: (a) \(\left[\mathrm{CO}_{2}\right]\) is increased? (b) The volume of the reaction vessel is decreased? (c) \(\mathrm{Ca}^{2+}(a q)\) is added? (d) \(\mathrm{CaCO}_{3}(s)\) is removed?

Suppose you have a reaction with many reactants. When you write the equilibrium expression for the reaction, do the reactant concentrations all go in the numerator or in the denominator? What mathematical operation(s) should be used for these concentrations?

An 8.00-L reaction vessel at \(491^{\circ} \mathrm{C}\) contains \(0.650\) mole of \(\mathrm{H}_{2}, 0.275\) mole of \(\mathrm{I}_{2}\), and \(3.00\) moles of HI. Assuming that the reaction is at equilibrium, determine the value of \(K_{\text {eq }}\) and comment on where the equilibrium lies. The reaction is: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)\)

Sparingly soluble calcium phosphate dissolves in water to yield an equilibrium calcium ion concentration of \(7.8 \times 10^{-6} \mathrm{M}\). (a) Write the balanced equilibrium equation for calcium phosphate dissolving in water. (b) Write the \(K_{\text {sp }}\) expression for calcium phosphate. (c) What is the equilibrium concentration of phosphate ion? (d) Calculate the value of \(K_{s p}\) for calcium phosphate (show your calculation).
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