Question

Suppose that a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4. What will be the concentration of H2SO4 in the battery after 3.50 A of current is drawn from the battery for 9.50 hours ?

please show all steps. I'm confused

Answer 1

Initial Molarity of H2SO4 is , n = Molarity x volume in L

= 5.00 M x 1.50 L

= 7.5 mol

The amount of charged , Q = (3.50A ) x (9.50 hrs) x ( 3600 s / hr) = 119700 C

So number of moles of electrons transferred = 119700 C /( 96500 C/mole of electrons )

   = 1.24 moles

Therefore the number of moles of H₂SO₄ consumed is = number of moles of electrons = 1.24 moles

So number of moles of H₂SO₄ left = initial moles - number of moles consumed

= 7.5 - 1.24

= 6.26 moles

So concentration of H₂SO₄ is = number of moles left / volume

= 6.26 mol / 1.50 L

= 4.17 M