Question

When solutions of aluminum sulfate and sodium hydroxide are mixed, a white gelatinous precipitate forms. (a) Write a balanced net ionic equation for the reaction. (b) What is the mass of the precipitate when \(2.76 \mathrm{~g}\) of aluminum sulfate in \(125 \mathrm{~mL}\) of solution is combined with \(85.0 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{NaOH}\) ? (c) What is the molarity of the ion in excess? (Ignore spectator ions and assume that volumes are additive.)

Short answer

a) The balanced net ionic equation for the reaction between aluminum sulfate and sodium hydroxide is: $$ \mathrm{2Al^{3+} + 6OH^- \rightarrow 2Al(OH)_3} $$ b) When 2.76 g of aluminum sulfate reacts with 85.0 mL of 0.2500 M sodium hydroxide solution, the mass of the precipitate formed is 0.376 g of aluminum hydroxide. c) The molarity of the excess hydroxide ions, ignoring spectator ions and assuming volumes are additive, is 0.0346 M.

Step-by-step solution

Write the balanced chemical equation

First, we need to identify the products of the chemical reaction. The reaction is between aluminum sulfate (\(\mathrm{Al_2(SO_4)_3}\)) and sodium hydroxide (\(\mathrm{NaOH}\)), producing aluminum hydroxide (\(\mathrm{Al(OH)_3}\)) and sodium sulfate (\(\mathrm{Na_2SO_4}\)). Here is the balanced chemical equation: $$ \mathrm{Al_2(SO_4)_3 + 6NaOH \rightarrow 2Al(OH)_3 + 3Na_2SO_4} $$

Write the balanced net ionic equation

Next, write the balanced net ionic equation by eliminating spectator ions. In this case, sodium (\(\mathrm{Na^+}\)) and sulfate (\(\mathrm{SO_4^{2-}}\)) are the spectator ions, as their presence does not change the outcome of the reaction. The net ionic equation is: $$ \mathrm{2Al^{3+} + 6OH^- \rightarrow 2Al(OH)_3} $$

Calculate the moles of aluminum sulfate and sodium hydroxide

We are given 2.76 g of aluminum sulfate, which we can convert to moles using the molar mass: $$ \mathrm{moles \: Al_2(SO_4)_3 = \frac{2.76\:g}{(2\times 26.98 + 3\times(4\times 16.00 + 32.07))\:g/mol} = 0.009658 \:mol} $$ We are also given 85.0 mL of 0.2500 M sodium hydroxide solution. Convert volume to liters and calculate the moles of sodium hydroxide: $$ \mathrm{moles \: NaOH = 0.2500 \frac{mol}{L} \times 0.085\:L = 0.02125\:mol} $$

Determine the limiting reactant

Using the balanced net ionic equation, 1 mole of aluminum ion reacts with 3 moles of hydroxide ions. Compare the mole ratios of aluminum ions and hydroxide ions: $$ \mathrm{\frac{moles\, Al^{3+}}{moles\, OH^-} = \frac{0.009658}{0.02125} \approx 0.454} $$ Since this ratio is less than the stoichiometric ratio (1:3), aluminum ions are the limiting reactant.

Calculate the mass of precipitate formed

Using the limiting reactant (aluminum ions), find the moles of aluminum hydroxide formed: $$ \mathrm{moles\: Al(OH)_3 = \frac{1}{2} \times moles\: Al_2(SO_4)_3} $$ $$ \mathrm{moles\: Al(OH)_3 = \frac{1}{2} \times 0.009658\:mol = 0.004829\:mol} $$ Then, calculate the mass of the precipitate: $$ \mathrm{mass\: Al(OH)_3 = moles\: Al(OH)_3 \times \frac{(26.98 + 3 \times 16.00)\:g/mol}{mol} = 0.004829 \:mol \times 78.00\:\frac{g}{mol} = 0.376 \:g} $$ The mass of the precipitate formed is 0.376 g of aluminum hydroxide.

Determine the concentration of excess ion

Since aluminum ions were the limiting reactant, sodium hydroxide is in excess, and the hydroxide ions are the excess ions. Calculate the moles of hydroxide ions remaining: $$ \mathrm{moles\: OH^-\: remaining = moles\: OH^- \:initial - 3 \times moles\: Al^{3+}} $$ $$ \mathrm{moles\: OH^-\: remaining = 0.02125 - (3 \times 0.009658) = 0.007276\:mol} $$ Finally, calculate the concentration of excess hydroxide ions, assuming additive volumes (total volume: 125 mL + 85.0 mL = 210.0 mL = 0.210 L): $$ \mathrm{Molarity\: OH^- = \frac{moles\: OH^-}{total\: volume\: L} = \frac{0.007276\:mol}{0.210\: L} = 0.0346\:M} $$ The molarity of the excess hydroxide ions is 0.0346 M.

Key concepts

Aluminum Hydroxide Precipitation

The precipitation of aluminum hydroxide is a classic example of a chemical reaction where an insoluble product, known as a precipitate, forms from soluble reactants. In this case, the reaction occurs between aluminum ions

• when aluminum sulfate dissolves, it releases aluminum ions into the solution
• and hydroxide ions, which come from sodium hydroxide solution.

As these ions combine, they form aluminum hydroxide, a gelatinous white solid that precipitates out of the solution.

Precipitation reactions are important in many fields, as they help in the removal or recovery of particular ions from a solution. For instance, water treatment processes often use precipitation to extract unwanted metals or compounds by converting them into a solid form that can be filtered out easily.

Limiting Reactants in Chemistry

In any chemical reaction, understanding the concept of limiting reactants is crucial. The limiting reactant is the substance that is completely consumed first in a chemical reaction. It determines the maximum amount of product that can be formed. In the given exercise, the limiting reactant is aluminum ion.

By using the ratio from the balanced chemical equation, we find that the ratio is less than required (1:3), indicating that aluminum ions will be used up first. Knowing the limiting reactant allows us to calculate how much product can be formed and how much of the other reactants remain unreacted (in excess).

• This information is important to predict the theoretical yield of a product in experiments.
• It also helps in optimizing industrial processes by ensuring that reactants are used efficiently.

Molarity Calculations

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. To calculate the molarity of a solution: 1. Determine the number of moles of solute present. 2. Measure the total volume of the solution in liters.

For example, in the original exercise, you calculated the moles of sodium hydroxide by multiplying its concentration (0.2500 M) with the volume in liters (0.085 L). To find the concentration of the excess hydroxide ions, the number of moles remaining was divided by the total volume of the combined solutions (0.210 L).

• Molarity is a fundamental concept in chemistry for preparing solutions with precise concentrations.
• It is also vital for stoichiometric calculations, particularly in determining the concentrations of ions present in excess after reactions.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. This process is accompanied by the breaking and forming of chemical bonds. In the example provided, aluminum sulfate and sodium hydroxide react to form aluminum hydroxide and sodium sulfate.

Balancing chemical equations is critical because it ensures that the same number of each type of atom appears on both sides of the equation, adhering to the law of conservation of mass. It's also important to understand net ionic equations, which provide clarity by indicating only the particles that participate directly in the chemical reaction.

• Such simplification helps in focusing on the actual changes occurring during the reaction and aids in eliminating unnecessary details from spectator ions.
• Understanding chemical reactions and equations also assists chemists in predicting product formation and understanding reaction dynamics.