# A 0.110 M H_2SO_4 solution is used to neutralize 10.0 mL of 0.085 M NaOH. What volume of the acid, in mL, is required to neutralize the base?

Jan 25, 2017

You need 3.86 mL of ${\text{H"_2"SO}}_{4}$ to neutralise the base.

#### Explanation:

So what we have to do is use stoichiometry to find the volume of acid needed to neutralise the base.

1: We need a balanced chemical equation:

$\text{H"_2"SO"_4(aq) + "2NaOH (aq)" rarr "Na"_2"SO"_4 (s) + "2H"_2"O} \left(l\right)$

So what we can find first is the moles of $\text{NaOH}$.

2: Solve for moles. Note that I changed 10 mL to 0.01 L.

${n}_{1} = c v$

$= \text{0.085 mol/L (0.010 L)}$

$= 8.5 \times {10}^{-} 4 \text{mol}$

3: Now we transfer the moles of $\text{NaOH}$ to $\text{H"_2"SO"_4}$.

To do that, we divide by $\text{NaOH}$'s coefficient, and multiply by ${\text{H"_2"SO}}_{4}$'s.

${n}_{2} = \frac{8.5 \times {10}^{-} 4 \text{mol}}{2}$

$= 4.25 \times {10}^{-} 4 \text{mol}$

There are $4.25 \times {10}^{-} 4$ mol of ${\text{H"_2"SO}}_{4}$.

4: With ${\text{H"_2"SO}}_{4}$'s moles and concentration (given), we can solve for volume.

$v = {n}_{\text{2}} / c$

$= \left(4.25 \times {10}^{-} 4 \text{mol")/("0.110 mol/L}\right)$

$\text{= 0.003,863,636}$ L = $\text{3.863,636}$ mL

Significant digits tells us to round the volume to 3 decimal places:

$3.86$ mL.

Therefore, you need $3.86$ mL of ${\text{H"_2"SO}}_{4}$ to neutralise the base.

Hope this helps :)