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Chapter 18: Problem 77

(a) Explain why \(\mathrm{Mg}(\mathrm{OH})_{2}\) precipitates when \(\mathrm{CO}_{3}{ }^{2-}\) ion is added to a solution containing \(\mathrm{Mg}^{2+}\) - (b) Will \(\mathrm{Mg}(\mathrm{OH})_{2}\) precipitate when \(4.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is added to \(1.00 \mathrm{~L}\) of a solution containing 125 Ppm of \(\mathrm{Mg}^{2+}\) ?

Short Answer

Expert verified
Mg(OH)鈧� precipitates when CO鈧兟测伝 is added to a solution containing Mg虏鈦� ions because it forms an insoluble compound that precipitates out of the solution. After calculating the ion concentrations of CO鈧兟测伝 and Mg虏鈦� in the given solution and comparing the ion product (IP) with the solubility product constant (Ksp) for Mg(OH)鈧�, we find that IP > Ksp. This indicates that the solution is supersaturated concerning Mg(OH)鈧�, and therefore, Mg(OH)鈧� will precipitate when 4.0 g of Na鈧侰O鈧� is added to 1.00 L of a solution containing 125 ppm of Mg虏鈦� ions.

Step by step solution

01

Discuss the chemical reaction

When CO鈧兟测伝 ions are added to a solution containing Mg虏鈦� ions, the following chemical reaction occurs: \(Mg^{2+} (aq) + 2OH^- (aq) \rightarrow Mg(OH)_{2} (s)\) This reaction forms Mg(OH)鈧�, which is insoluble in water and precipitates out of the solution.
02

Calculate ion concentrations

We'll calculate the concentrations of CO鈧兟测伝 and Mg虏鈦� ions in the solution after adding 4.0 g of Na鈧侰O鈧� to 1.00 L of the solution. First, let's find the concentration of CO鈧兟测伝 ions: - We have 4.0 g of Na鈧侰O鈧�. - The molar mass of Na鈧侰O鈧� is \(2\times(22.99)+12.01+3\times(16.00) = 105.99~g/mol\). - Therefore, the moles of Na鈧侰O鈧� = \(\frac{4.0~g}{105.99~g/mol} = 0.0377~mol\). - The concentration of CO鈧兟测伝 ions is \(\frac{0.0377~mol}{1.00~L} = 0.0377~M\). Now, let's calculate the concentration of Mg虏鈦� ions: - We have 125 ppm of Mg虏鈦� ions, which means there are 125 mg of Mg虏鈦� ions per 1 L of the solution. - The molar mass of Mg is 24.30 g/mol. - Therefore, the moles of Mg虏鈦� = \(\frac{125~mg}{24.30~g/mol} \times \frac{1~g}{1000~mg} = 0.00514~mol\). - The concentration of Mg虏鈦� ions is \(\frac{0.00514~mol}{1.00~L} = 0.00514~M\).
03

Determine if Mg(OH)鈧� will precipitate

Next, let's determine the ion product and compare it with the solubility product constant, \(K_{sp}\), of Mg(OH)鈧�. The ion product, IP, is given by the expression: \[IP = [Mg^{2+}][OH^-]^2\] In this case, the hydroxide ions come from the dissociation of CO鈧兟测伝 ions. We know that for each CO鈧兟测伝 ion that dissolves, one OH鈦� ion is formed. So, the concentration of OH鈦� ions is equal to the concentration of CO鈧兟测伝 ions. Hence, \[[OH鈦籡 = [CO鈧僞{2-}] = 0.0377~M\] Now, calculate the ion product: \[IP = (0.00514)(0.0377)^2 = 7.13 \times 10^{-6}\] The solubility product constant, \(K_{sp}\), for Mg(OH)鈧� is given as \(K_{sp} = 5.61 \times 10^{-12}\). Comparing the ion product and the solubility product constant, we see that IP > \(K_{sp}\). That means, the solution is supersaturated with respect to Mg(OH)鈧�, and Mg(OH)鈧� will precipitate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant
The solubility product constant, represented as \( K_{sp} \), is a valuable tool in chemistry used to determine the solubility of a compound in a solution. It describes the equilibrium point for the dissolution of a sparingly soluble compound. For the compound magnesium hydroxide, \( \text{Mg(OH)}_2 \), this constant captures how much of the compound can dissolve in water to form its ions Mg虏鈦� and OH鈦�. The expression for the solubility product constant for magnesium hydroxide is:
  • \( K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \)
This equation indicates that the solubility of a compound is dependent on the concentrations of its ions in solution. A situation where the ion product exceeds \( K_{sp} \) results in precipitation, as the solution becomes supersaturated. Hence, by comparing the calculated ion product to the \( K_{sp} \), we can predict whether a precipitate will form. This is the key method used to determine the formation of magnesium hydroxide precipitate in chemical solutions.
Magnesium Hydroxide
Magnesium hydroxide, \( \text{Mg(OH)}_2 \), is a white solid known for its low solubility in water. When a solution containing magnesium ions \( \text{(Mg}^{2+}) \) meets a source of hydroxide ions \( \text{(OH}^-) \), the conditions are ripe for forming magnesium hydroxide. This compound appears as a precipitate because it does not dissolve easily at room temperature.
Magnesium hydroxide is often encountered in antacids and laxatives and is a simple but essential material in the real world. Its formation fascinates chemists as it considers ion interactions in a reaction. Understanding these concepts ensures that the manipulation of solutions in laboratory settings is accurate, which could be critical in industrial applications and academic pursuits.
Ion Concentration Calculation
Calculating ion concentrations in a solution involves stoichiometry and a grasp of molecular masses and conversions. When you add a compound like sodium carbonate, \( \text{Na}_2\text{CO}_3 \), into a solution, its dissociation provides concentrations of specific ions鈥攊n this case, \( \text{CO}_3^{2-} \) and a resultant impact on \( \text{OH}^- \) concentration.
  • The process starts by determining the moles of \( \text{Na}_2\text{CO}_3 \) by dividing the mass of the substance by its molar mass.
  • The concentration of ions is then calculated by dividing the moles by the volume of the solution in liters.
This thorough calculation helps predict whether a reaction between ions will result in a precipitate. For magnesium hydroxide, understanding the ion concentrations post-reaction is crucial. This guides us in assessing whether conditions allow for precipitation based on the comparison with the \( K_{sp} \). Accurate ion concentration calculation ensures precise interpretations of chemical behavior in solutions.

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Most popular questions from this chapter

A friend of yours has seen each of the following items in newspaper articles and would like an explanation: (a) acid rain, (b) greenhouse gas, (c) photochemical smog, (d) ozone depletion. Give a brief explanation of each term and identify one or two of the chemicals associated with each. s in this respect?

(a) With respect to absorption of radiant energy, what distinguishes a greenhouse gas from a non-greenhouse gas? (b) \(\mathrm{CH}_{4}\) is a greenhouse gas, but \(\mathrm{N}_{2}\) is not. How might the molecular structure of \(\mathrm{CH}_{4}\) explain why it is a greenhouse gas?

(a) Which of the following ionic species could be responsible for hardness in a water supply: \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Mg}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Na}^{+}\)? (b) What properties of an ion determine whether it will contribute to water hardness?

(a) What is the difference between a CFC and an HFC? (b) It is estimated that the lifetime for HFCs in the stratosphere is 2-7 years. Why is this number significant? (c)Why have HFCs been used to replace CFCs? (d) What is the major disadvantage of HFCs as replacements for CFCs?

In the following three instances which choice is greener in a chemical process? Explain. (a) A reaction that can be run at \(350 \mathrm{~K}\) for \(12 \mathrm{~h}\) without a catalyst or one that can be run at \(300 \mathrm{~K}\) for \(1 \mathrm{~h}\) with a reusable catalyst. (b) A reagent for the reaction that can be obtained from corn husks or one that is obtained from petroleum. (c) A process that produces no by-products or one in which the by-products are recycled for another process.
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