Let a fully charged lead storage battery contains 1.5 l of 5m h2so4 what will be the concentration of h2so4 in the battery after 2.5 ampere current is drawn from the battery for 6 hour ?
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(2.50 A ) (8.50 hrs) (3600 s/hr) = 30,600 C
number of moles of electrons = 30,600 C / 96,500 C/mole of e- = 0.317 mole
0.317 mole of electrons --> 0.317 moles of H2SO4 consumed
Initial moles of sulfuric acid = 1.50 L x 5.00M = 7.50 moles
Final moles of sulfuric acid = 7.50 - 0.317 = 7.18 moles H2SO4
Final concentration of sulfuric acid = 7.18 moles H2SO4 / 1.50L = 4.79 M H2SO4
number of moles of electrons = 30,600 C / 96,500 C/mole of e- = 0.317 mole
0.317 mole of electrons --> 0.317 moles of H2SO4 consumed
Initial moles of sulfuric acid = 1.50 L x 5.00M = 7.50 moles
Final moles of sulfuric acid = 7.50 - 0.317 = 7.18 moles H2SO4
Final concentration of sulfuric acid = 7.18 moles H2SO4 / 1.50L = 4.79 M H2SO4
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