A 65 mL sample of a solution of sulfuric acid is neutralized by 46 mL of a 0.14 M sodium hydroxide solution. Calculate the molarity of the sulfuric acid solution.

Answer in units of mol/L

To calculate the molarity of the sulfuric acid solution, you need to use the concept of molar ratios and the formula for molarity.

The balanced chemical equation for the neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is:

H2SO4 + 2NaOH -> Na2SO4 + 2H2O

From the equation, you can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide. Therefore, the molar ratio is 1:2.

First, we need to determine the number of moles of sodium hydroxide used in the reaction. This can be calculated using the formula:

moles = volume (in liters) × molarity

Given that the volume of the sodium hydroxide solution is 46 mL or 0.046 L, and the molarity is 0.14 M, we can substitute these values into the formula:

moles of NaOH = 0.046 L × 0.14 M
moles of NaOH = 0.00644 moles

Since the molar ratio between sulfuric acid and sodium hydroxide is 1:2, the number of moles of sulfuric acid used in the reaction is also 0.00644 moles.

Now, we can calculate the molarity of the sulfuric acid solution using the formula for molarity:

Molarity = moles / volume (in liters)

Given that the volume of the sulfuric acid solution is 65 mL or 0.065 L, we can substitute these values into the formula:

Molarity of sulfuric acid = 0.00644 moles / 0.065 L
Molarity of sulfuric acid = 0.099 mol/L

Therefore, the molarity of the sulfuric acid solution is 0.099 M.

You should be able to do this with the NaOH vs HCl problem. Post your work if you get stuck.