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Calculate the molarity of the original sample of H2SO4 solution.?
3. A 0.500-L sample of H2SO4 solution was analyzed by taking an aliquot of 100.0 mL and reacting it with 50.0 mL of 0.213 M NaOH(aq). After the reaction occurred, the excess base required titration against 13.21 mL of 0.103 M HCl(aq) for complete neutralization.
Calculate the molarity of the original sample of H2SO4 solution.
1 Answer
- hcbiochemLv 75 years agoFavorite Answer
Work the problem backward:
In the final titration, moles HCl required = 0.01321 L X 0.103 mol/L = 1.36X10^-3 mol HCl. So, that solution still contained 1.36X10^-3 mol NaOH
But, moles NaOH added originally = 0.0500 L X 0.213 mol/L = 0.0106 mol NaOH
So, the moles NaOH consumed by reaction with H2SO4 was: 0.0106 mol - 0.00136 mol = 9.29X10^-3 mol NaOH
So, since H2SO4 requires 2 moles NaOH to neutralize each mole,
moles H2SO4 = 9.29X10^-3 mol NaOH X (1 mol H2SO4 / 2 mol NaOH) = 4.64X10^-3 mol
[H2SO4] = 4.64X10^-3 mol H2SO4 / 0.500 L = 9.29X10^-3 M H2SO4
