# 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...

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!

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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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