If you have 160.1 mL of a 1.025 M solution of sodium hydroxide, how many milliliters of 0.550 M solution of a sulfuric acid do you need in order to neutralize it? ___ mL

Answer 1

Balanced chemical reaction:

2NaOH (aq ) + H2SO4 ( aq ) -----------> Na2SO4 (aq ) + 2H2O (l)

For NaOH:

M1 = 1.025 M/L

V1 = 160.1 ml

For H2SO4:

M2 = 0.550 M/L

V2 = ?

Applying Molarity Equation:

M1V1 = M2V2

------------ ---------

2 1

1.025M/L * 160.1 ml = 0.550M/L * V2 ml

------------------- -----------------

2 1

1.025M/L * 160.1ml * 1

V2 = ---------------------------- = 149.18 ml

0.550 M/L * 2

Answer 2

Hi Madison! First, we should always write the neutralization reaction and balance it:

2NaOH + H2SO4 —> 2H2O + Na2SO4

Notice now it takes 2 mols of NaOH to neutralize 1 mol of H2SO4. So we can determine the mols of NaOH we have from the given information:

160.1 mL x 1L/1000 mL x 1.025 mols NaOH/ L = 0.1641 mol NaOH

Using the stoichiometric relationship, let’s determine the amount of mols of H2SO4 needed to neutralize 0.1641 mols of NaOH:

0.1641 mols NaOH x 1 mol H2SO4/ 2 mols NaOH = 0.0820 mols H2SO4

Since we know the concentration of H2SO4 and the mols needed, we can determine the volume needed in mL:

0.08205 mols H2SO4 x 1 L/ 0.550 mols H2SO4 * 1000 mL/ 1 L = 149.2 mL H2SO4

Since there are 3 sig figs in the concentration, and 4 sig figs in the mols, the volume must be 3 sig figs, so this rounds to 149 mL H2SO4

I hope this helps!