Question

What would the resulting pH of a solution obtained by mixing 35mL of 0.30 M H2SO4 with 10mL of 0.25 M NaOH solution?

Answer 1

mmoles of H2SO4 = 0.3M x 35 mL = 10.5 mmol

mmoles of NaOH = 0.25M x 10 mL = 2.5mmol

let see the balanced equation

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

each mole of H2SO4 required 2 moles of NaOH accordingly

2.5 mmol of NaOH required 2.5 / 2 = 1.25 mmol of H2SO4

mmoles of H2SO4 remaining = 10.5-1.25 = 9.25 mmol

total volume = 35+10 = 45 mL

concentration of H2SO4 remaining = 9.25 mmol / 45 mL   = 0.2055 M

H2SO4 is a diprotic acid so concentration of H2SO4 = 2 x concentration of H+

= 2 x 0.2055 M

= 0.41 M

pH = -logH+ = -log(0.41) = 0.38