Question

If 85.00 mL of 0.100 M HCl is mixed with 25.00 mL of 0.200 M H2SO4 and 50.00 mL of 0.400 M NaOH. What is the resulting pH of the final solution?

Please show all work and how the formulas are derived if modified.

Answer 1

Moles = M x V ( in Liters) is formula

Moles of HCl = 0.1 x 0.085 = 0.0085

Moles of H+ from HCl = 0.0085   ( since 1HCl gives 1H+ per dissociation)

Moles of H2SO4 = 0.2 x 0.025 = 0.005

Moles of H+ = 2 x 0.005 = 0.01        ( since 1H2SO4 gives 2H+ per dissociation)

Total H+ moles from HCl and H2SO4 = 0.0085+0.01 =0.0185

Moles of NaOH = 0.4 x 0.05 = 0.02

Moles of OH- = 0.02 ( since 1NaOH gves 1OH-)

H+ and OH- reatc in 1:1 ratio to form H2O

Moles of OH- left after reacting with 0.0185 moles H+ = 0.02-0.0185 = 0.0015

solution volume = 0.085 + 0.025 + 0.05 = 0.16 Litters

[OH-] = moles / volume = 0.0015 / 0.16 = 0.0093375

pOH = -log [OH-] = -log ( 0.009375) = 2

pH = 14 - poH = 14-2= 12