A sample of battery acid is be analyzed for its sulfuric acid content

sulfuric-acid
acid-analyzed

#1

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 %