42.55mL of a standardized 1.005 M NaOH solution was necessary to exactly neutralize 30.0mL of an aqueous H2SO4 solution of unknown concentration. What is the molarity of the H2SO4 solution?
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
mols NaOH - M x L = ?
Using the coefficients in the balanced equation, convert mols NaOH to mols H2SO4.
Then M H2SO4 = mols H2SO4/L H2SO4
To find the molarity of the H2SO4 solution, we can use the concept of stoichiometry and the equation for the neutralization reaction between NaOH and H2SO4.
The balanced equation for the reaction is:
2 NaOH + H2SO4 -> Na2SO4 + 2 H2O
From the equation, we can see that it takes two moles of NaOH to react with one mole of H2SO4.
Given that 42.55 mL of a 1.005 M NaOH solution was used to neutralize 30.0 mL of the H2SO4 solution, we can use the following equation (derived from the stoichiometry of the reaction) to calculate the molarity of the H2SO4 solution:
Molarity of NaOH solution * Volume of NaOH solution = Molarity of H2SO4 solution * Volume of H2SO4 solution
(1.005 M) * (42.55 mL) = Molarity of H2SO4 solution * (30.0 mL)
Let's solve for the molarity of the H2SO4 solution:
Molarity of H2SO4 solution = (1.005 M * 42.55 mL) / 30.0 mL
Molarity of H2SO4 solution = 1.427475 M
Therefore, the molarity of the H2SO4 solution is approximately 1.43 M.