Question

Given the following balanced chemical reaction, what volume of 3.0M H2SO4 is required to neutralize 60.0 mL of 0.5M NaOH? Be sure to include the formula and show your work for each step in the calculation.

Answer 1

Answer:

$\boxed{\text{10 mL}}$

Explanation:

1. Write the balanced chemical equation.

$\text{2NaOH} + \text{H _{2} SO _{4} } \longrightarrow\ \text{Na _{2} SO _{4} } + 2\text{H{ _{2} O}}$

2. Calculate the moles of NaOH

$\text{Moles of NaOH} =\text{60.0 mL NaOH} \times \dfrac{\text{0.5 mmol NaOH}}{\text{1 mL NaOH}} = \text{30 mmol NaOH}$

3. Calculate the moles of H₂SO₄.

$\text{Moles of H _{2} SO _{4} }=\text{30 mmol NaOH} \times \dfrac{\text{1 mmol H _{2} SO _{4} } }{\text{2 mmol NaOH}} = \text{30 mmol H _{2} SO _{4} }$

4. Calculate the volume of H₂SO₄

$c = \text{30 mmol H _{2} SO _{4} } \times \dfrac{\text{1 mL H _{2} SO _{4} }}{\text{3.0 mmol H _{2} SO _{4} }} = \text{10 mL H _{2} SO _{4} }$

The titration will require $\textbf{10 mL}$ H₂SO₄.