Question

A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.695 M HCl and 0.335 M H₂SO₄. What volume of 0.62 M NaOH would be required to completely neutralize all of the acid in 772.7 mL of this solution?

Answer 1

    2NaOH + H2so4 ---> Na2SO4 + 2H2O

NaOH + HCl    ----> NaCl + H2o

from the mixture = 3 mole H+ present ( 1 mole H2SO4 = 2MOLE H+ ,

HCL = 1 MOLE H+)

IF total acid dissocitae

No of moles of HCl = 0.695*Vin L

= 0.695*0.7727 = 0.537 mole

No of mole of H2so4 = 0.335*0.7727 = 0.2588 mole

total H+ ions = 0.537 + (2*0.2588)

    = 1.0546 mole

No of moles of OH- = required = 1.0546 mole

No of moles of NaOH = 1.0546 mole

volume of NaOH required V = 0.62/1.0546 = 0.5879 L

   = 587.9 ml