Question

What volume of 12.7 mol/L H2SO4(aq) solution is required to neutralize 5.00 L of 0.100 mol/L NaOH(aq) solution?

Answer 1

Given:

$$V_{NaOH} = 5.00\ L$$ $$M_{H_2 SO_4} = 12.7 \ mol/L$$ $$M_{NaOH}= 0.100 \ mol/L$$

Solution:

$$H_2 SO_4 + 2NaOH = Na_2 SO_4 + 2H_2 O$$

Clearly there is a 1:2 equivalence, so

$$n_{H_2 SO_4} = \frac{1}{2}* n_{NaOH}$$

and as a first step we calculate the number of moles of sulfuric acid:

$$n_{NaOH} = M_{NaOH} * V_{NaOH}$$ $$n_{NaOH} = 0.100 mol/L * 5L = 0.5mol$$

from first formula:

$$n_{H_2 SO_4} = 0.5* n_{NaOH} = 0.5* 0.5 mol = 0.25 mol$$

From amount of substance we can get volume:

$$V_{H_2SO_4} = \frac{n_{H_2SO_4}}{M_{H_2SO_4}} = \frac{0.25 mol}{12.7 mol/L} \approx 0.2L$$

Answer:

$$V_{H_2SO_4} \approx 0.2 L$$