Question

If you have 190.1 mL of a 0.6650 M solution of sodium hydroxide, how many mL of a 0.950 M solution of sulfuric acid do you need in order to neutralize it?

Answer 1

To determine the volume of the sulfuric acid solution needed to neutralize the sodium hydroxide solution, we can use the concept of stoichiometry and the balanced chemical equation between sodium hydroxide (NaOH) and sulfuric acid (H2SO4).

The balanced equation for the neutralization reaction is:

2 NaOH + H2SO4 -> Na2SO4 + 2 H2O

From the equation, we can see that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid. Therefore, the molar ratio is 2:1.

Given:

Volume of sodium hydroxide solution = 190.1 mL

Molarity of sodium hydroxide solution = 0.6650 M

Molarity of sulfuric acid solution = 0.950 M

To find the volume of the sulfuric acid solution required, we can use the following equation:

(Volume of sodium hydroxide solution) × (Molarity of sodium hydroxide solution) × (Molar ratio) = (Volume of sulfuric acid solution) × (Molarity of sulfuric acid solution)

Let's solve for the volume of the sulfuric acid solution:

(190.1 mL) × (0.6650 M) × (1/2) = (Volume of sulfuric acid solution) × (0.950 M)

Volume of sulfuric acid solution = (190.1 mL) × (0.6650 M) × (1/2) / (0.950 M)

Volume of sulfuric acid solution ≈ 139.94 mL

Therefore, approximately 139.94 mL of the 0.950 M sulfuric acid solution is needed to neutralize 190.1 mL of the 0.6650 M sodium hydroxide solution.