# What volume of 0.125 mol/L NaOH (aq) is required to react completely with 15.0 mL of 0.100 mol/L Al2(SO4)3 (aq)? If anyone could PLEASE not only the answer the question, but explain the steps and why these steps are used. Thanks! The key reaction happening here is the aluminum cation reacting with the hydroxide anion to produce aluminum hydroxide.  The equation is shown below.

Al3+ + 3OH- --> Al(OH)3

Aluminum hydroxide will precipitate from aqueous solution.  The key here is that 3 moles of hydroxide are required to completely react with...

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The key reaction happening here is the aluminum cation reacting with the hydroxide anion to produce aluminum hydroxide.  The equation is shown below.

Al3+ + 3OH- --> Al(OH)3

Aluminum hydroxide will precipitate from aqueous solution.  The key here is that 3 moles of hydroxide are required to completely react with each 1 mole of aluminum.  So first we need to figure out how many moles of Al3+ we are dealing with here.   If we have 15 mL of a 0.1 M solution of Al2(SO4)3:

0.1 mole/L * 0.015 L = 0.0015 moles of Al2(SO4)3

Each mole of Al2(SO4)3 produces 2 moles of Al3+:

0.0015 moles Al2(SO4)3 * 2 = 0.003 moles Al3+

Since each mole of Al3+ requires 3 moles of OH- to react with it we should multiply by 3 to get the number of moles of OH- required:

0.003 moles Al3+ * (3 moles OH-/1 mole Al3+) = 0.009 moles OH-

Now we need to figure out the volume of 0.125 M solution of NaOH needed to get 0.009 moles OH-.

0.009 moles OH- * (1 L/0.125 moles) = 0.072 L = 72 mL of solution

So 72 mL of the NaOH solution is required.

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