A 69 mL sample of a solution of sulfuric acid is neutralized by 41 mL of a .161 M sodium hydroxide solution. Calculate the molarity of the sulfuric acid solution. Answer in units of mol/L.
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
mols NaOH = M x L = ?
mols H2SO4 = 1/2 x mols NaOH (Look at the coefficients in the balanced equation.)
Then M H2SO4 = mols H2SO4/L H2SO4 = ?
To calculate the molarity of the sulfuric acid solution, we can use the concept of stoichiometry and the equation:
acid + base → salt + water
In this case, the balanced equation for the neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the equation, we can see that the ratio between H2SO4 and NaOH is 1:2. Therefore, one mole of sulfuric acid reacts with two moles of sodium hydroxide.
Given the volume of sodium hydroxide solution used (41 mL) and its concentration (0.161 M), we can calculate the number of moles of sodium hydroxide used:
moles of NaOH = volume (L) × concentration (M)
= 0.041 L × 0.161 mol/L
= 0.006601 mol
Due to the stoichiometry relationship, we know that the number of moles of sulfuric acid is half the number of moles of sodium hydroxide used:
moles of H2SO4 = 0.006601 mol ÷ 2
= 0.0033005 mol
Finally, we need to calculate the molarity of the sulfuric acid solution using the volume of the solution (69 mL = 0.069 L) and the number of moles of sulfuric acid calculated:
molarity (M) = moles ÷ volume (L)
= 0.0033005 mol ÷ 0.069 L
≈ 0.04783 mol/L
Therefore, the molarity of the sulfuric acid solution is approximately 0.04783 mol/L.