Question

What is the pH of 70.0 mL of a 0.50 M HCl solution after the following additions? 0.00 mL of 0.50 M NaOH, 30.0 mL of 0.50 M NaOH, 70.0 mL of 0.50 M NaOH, 80.0 mL of 0.50 M NaOH

Answer 1

HCl + NaOH -------> NaCl + H2O

millimoles of HCl = 70 x 0.5 = 35.0

1) initially only HCl present

[HCl] = [H⁺] = 0.50M

pH = - log [H⁺]

pH = - log [0.50]

pH = 0.30

2) millimoles of NaOH added = 30 x 0.50 = 15

35 - 15 = 20 millimoles of HCl left

[HCl] = 20 / 70+30 = 0.2 M

pH = - log [0.20]

pH = 0.70

3) millimoles of NaOH added = 70 x 0.5 = 35

35 - 35 = all HCl becomes NaCl

as NaCl is neutral salt

pH = 7.0

4) millimoles of NaOH = 80 x 0.5 = 40

40 - 35 = 5.0 millimoles NaOH left

[NaOH] = 5.0 / 70+80 = 0.033 M

as NaoH is strong base  

[NaOH] = [OH^-] = 0.033 M

pOH = - log [OH^-]

pOH = - log [0.033]

pOH = 1.48

pH = 14 - 1.48

pH = 12.52