Chapter 17: Problem 101
Predict whether a precipitate forms if you mix 75.0 mL of a NaOH solution with pOH = 2.58 with 125.0 mL of a 0.018 M MgCl2 solution. Identify the precipitate, if any.
Short Answer
Expert verified
Yes, a precipitate will form when NaOH is mixed with MgCl2, and the precipitate is magnesium hydroxide (Mg(OH)₂).
Step by step solution
01
Calculate the hydroxide ion concentration [OH⁻]
Use the pOH value to find the hydroxide ion concentration using the formula: \( [OH^-] = 10^{-pOH} \). Plug in the given pOH value to get \( [OH^-] = 10^{-2.58} = 2.63 \times 10^{-3} \) M.
02
Calculate the moles of hydroxide ions
Multiply the concentration of hydroxide ions by the volume of the NaOH solution in liters to obtain the moles of hydroxide ions. \( Moles_{OH^-} = [OH^-] \times Volume_{NaOH}(L) = 2.63 \times 10^{-3} \times 0.075 L = 1.97 \times 10^{-4} \) moles.
03
Calculate the moles of Mg²⁺ ions
Multiply the concentration of the MgCl₂ solution by its volume in liters to obtain moles of Mg²⁺. \( Moles_{Mg^{2+}} = 0.018 M \times 0.125 L \) = \( 2.25 \times 10^{-3} \) moles.
04
Determine the limiting reactant
Compare the molar ratio of Mg²⁺ and OH⁻ required to form Mg(OH)₂. Since the reaction between Mg²⁺ and OH⁻ to form Mg(OH)₂ has a stoichiometry of 1:2, compare twice the moles of Mg²⁺ to the moles of OH⁻. The limiting reactant is the one in the least molar amount considering the stoichiometry.
05
Predict the formation of a precipitate
If there are enough moles of OH⁻ to react with all the available Mg²⁺ ions (considering the stoichiometry), then a precipitate of Mg(OH)₂ will form. Since \( 2 \times 2.25 \times 10^{-3} \) moles of OH⁻ are required and there are \( 1.97 \times 10^{-4} \) moles of OH⁻ available, there are not enough OH⁻ ions to precipitate all the Mg²⁺; hence a precipitate of Mg(OH)₂ will form.
06
Identify the precipitate
The precipitate formed in this reaction is magnesium hydroxide, Mg(OH)₂, due to the reaction between Mg²⁺ and OH⁻.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hydroxide Ion Concentration
Understanding the hydroxide ion concentration is key when predicting the outcomes of reactions in chemistry. Hydroxide ions (OH^-), are crucial when it comes to aqueous solutions, especially those with basic or alkaline properties. The concentration of these ions is often derived from the pOH value, which is the negative logarithm of the hydroxide ion concentration. This is similarly related to pH, but focuses on OH^- instead of H^+.
In our exercise, the pOH of a solution of sodium hydroxide (NaOH) is given as 2.58. To find the hydroxide ion concentration, we use the equation [OH^-] = 10^{-pOH}. Thus, the concentration of hydroxide ions can be calculated leading to a better understanding of the solution's basicity. This calculation is the foundational step for predicting whether a reaction will occur and if a precipitate will form in our given example.
In our exercise, the pOH of a solution of sodium hydroxide (NaOH) is given as 2.58. To find the hydroxide ion concentration, we use the equation [OH^-] = 10^{-pOH}. Thus, the concentration of hydroxide ions can be calculated leading to a better understanding of the solution's basicity. This calculation is the foundational step for predicting whether a reaction will occur and if a precipitate will form in our given example.
Limiting Reactant
The concept of the limiting reactant is pivotal in predicting the extent of a chemical reaction. In any chemical reaction, reactants are consumed to form products. However, not all reactants may be present in the correct stoichiometric ratios; the reactant that is used up first and limits the amount of product that can be formed is known as the limiting reactant.
In the context of the provided example, once the moles of reactants, hydroxide ions (OH^-), and magnesium ions (Mg^{2+}), have been calculated by using their respective concentrations and volumes, determining the limiting reactant will tell us whether a precipitate will form. The ratio of reactants needed according to the balanced chemical equation (stoichiometry) for Mg(OH)₂ establishes which reactant will be exhausted first, thereby ceasing the reaction.
In the context of the provided example, once the moles of reactants, hydroxide ions (OH^-), and magnesium ions (Mg^{2+}), have been calculated by using their respective concentrations and volumes, determining the limiting reactant will tell us whether a precipitate will form. The ratio of reactants needed according to the balanced chemical equation (stoichiometry) for Mg(OH)₂ establishes which reactant will be exhausted first, thereby ceasing the reaction.
Stoichiometry
Stoichiometry is a section of chemistry that involves the calculation of relative quantities of reactants and products in chemical reactions. It is based on the conservation of mass where the total mass of reactants equals the total mass of products. Stoichiometry hinges on the balanced chemical equation and the mole concept.
In relation to the reaction between Mg^{2+} and OH^-, the stoichiometric coefficients dictate that one mole of magnesium ions reacts with two moles of hydroxide ions to form one mole of magnesium hydroxide (Mg(OH)₂). This 1:2 ratio is important when it comes to determining the limiting reactant and predicting the amount of precipitate produced during the reaction.
In relation to the reaction between Mg^{2+} and OH^-, the stoichiometric coefficients dictate that one mole of magnesium ions reacts with two moles of hydroxide ions to form one mole of magnesium hydroxide (Mg(OH)₂). This 1:2 ratio is important when it comes to determining the limiting reactant and predicting the amount of precipitate produced during the reaction.
Precipitate Identification
Precipitate identification involves determining the solid product that forms when two aqueous solutions react together. This solid, called a precipitate, is typically an insoluble ionic compound. Predicting whether a precipitate will form can involve calculations that are rooted in stoichiometry and limiting reactant concepts as well as solubility rules.
When mixtures of different ions in solution react together, insoluble compounds can form if the product falls beyond its solubility product (K_{sp}). In our exercise, the reaction between Mg^{2+} ions and OH^- ions leads to the formation of magnesium hydroxide, Mg(OH)₂, a compound with low solubility in water and thus forms a precipitate under the right conditions. Identifying precipitates is crucial in both laboratory chemistry and industrial applications.
When mixtures of different ions in solution react together, insoluble compounds can form if the product falls beyond its solubility product (K_{sp}). In our exercise, the reaction between Mg^{2+} ions and OH^- ions leads to the formation of magnesium hydroxide, Mg(OH)₂, a compound with low solubility in water and thus forms a precipitate under the right conditions. Identifying precipitates is crucial in both laboratory chemistry and industrial applications.