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Chapter 4: Problem 26

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

Expert verified
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

01

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: $$ \mathrm{Al_2(SO_4)_3 + 6NaOH \rightarrow 2Al(OH)_3 + 3Na_2SO_4} $$
02

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: $$ \mathrm{2Al^{3+} + 6OH^- \rightarrow 2Al(OH)_3} $$
03

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: $$ \mathrm{moles \: NaOH = 0.2500 \frac{mol}{L} \times 0.085\:L = 0.02125\:mol} $$
04

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.
05

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.
06

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.

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