Problem 70 - Chemistry and Chemical Reactivity

Question

Will a precipitate of \(\mathrm{Mg}(\mathrm{OH})_{2}\) form when \(25.0 \mathrm{mL}\) of 0.010 M NaOH is combined with 75.0 mL of a 0.10 M solution of magnesium chloride?

Step-by-Step Solution

Answer

Precipitate of Mg(OH)₂ will not form.

Step 1: Write the Reaction Equation

Combine the NaOH and MgCl₂ to form the possible precipitate and products. The balanced chemical equation is:\[\text{MgCl}_2 + 2\text{NaOH} \rightarrow \text{Mg(OH)}_2 \downarrow + 2\text{NaCl}.\]

Key Concepts

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions are combined, and an insoluble solid forms. This solid is known as the precipitate. These reactions are a type of double displacement reaction, where parts of two compounds switch places to form two new compounds. In the example given, when sodium hydroxide (NaOH) is mixed with magnesium chloride (MgCl₂), magnesium hydroxide (Mg(OH)₂) can potentially precipitate out, and sodium chloride (NaCl) remains dissolved in the solution. Precipitation reactions can be predicted by using solubility rules, which outline which compounds are soluble and which are not. Solubility rules indicate that most chloride salts are soluble, except for those of silver, lead, and mercury. Hydroxides, however, tend to be less soluble, especially those of metals like magnesium. This makes the formation of magnesium hydroxide a prime candidate for precipitation.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using balanced chemical equations to quantitatively understand how much of each substance is used and produced. For the reaction between NaOH and MgCl₂, the balanced equation tells us that two moles of NaOH react with one mole of MgCl₂ to form one mole of Mg(OH)₂ and two moles of NaCl. This 2:1 ratio is crucial for determining how much of each reactant is needed and whether a precipitate will form.In the scenario provided, we must calculate the number of moles of each reactant present in the solutions. Using the formula \( ext{moles} = ext{concentration} \times ext{volume} \) (with volume in liters), we calculate the moles of NaOH and MgCl₂. From 25.0 mL of 0.010 M NaOH, there are 0.00025 moles of NaOH. From 75.0 mL of 0.10 M MgCl₂, there are 0.0075 moles of MgCl₂. This helps us understand the quantity of Mg(OH)₂ that could form.

Solubility Product

The solubility product constant, symbolized as Ksp, is an expression that describes the extent to which a compound can dissolve in water. It is crucial in predicting the formation of a precipitate in a chemical reaction. The Ksp value is unique for each compound and indicates whether a precipitate will form under given conditions.For magnesium hydroxide, \( ext{Mg(OH)}_2 \), the Ksp expression is given by \( ext{Ksp} = [ ext{Mg}^{2+}][ ext{OH}^-]^2 \). By comparing the ion product (the calculated equivalent of \( ext{Ksp} \) based on current concentrations) to the actual \( ext{Ksp} \) value from literature, one determines if precipitation will occur. If the ion product exceeds the Ksp, the solution is supersaturated, leading Mg(OH)₂ to precipitate. Conversely, if the ion product is below the Ksp, then no precipitate forms, and the solution remains unsaturated.