A sample of battery acid is be analyzed for its sulfuric acid content . A 1.00 ml sample (that weighs 1.239g) is diluted to 250.0 ml and 10.00 ml of this diluted solution requires 32.44 mL of 0.00498 M Ba(OH)2 for its titration. What is the mass percent of H2SO4 in the battery acid? ( Assume that comlete ionization and neutralization of the H2SO4 occurs).
Answer:
Ba(OH)2 + H2SO4 => BaSO4 + 2H2O
moles of base at the eq. point of the titration = .00498* 32.44/ 1000
0.000161 moles Ba(OH)2 = moles of the acid at the eq point
these are in the 10 ml aliquot
so total moles in the 250 ml flask = 0.000161 * 250/10
=0.004025 moles of H2SO4
these are the number of moles in 1 ml of the battery acid
= .004025 * 98 g of H2SO4
= .3944 g
mass % of H2SO4 in the sample =
= .3944 / 1.239
= 31.83 %