Question

Calculate the pH of the solution resulting from the addition of 30.0 mL of 0.200 M HClO4 to 60.0 mL of 0.150 M NaOH

Answer 1

we have:

Molarity of HClO4 = 0.2 M

Volume of HClO4 = 30 mL

Molarity of NaOH = 0.15 M

Volume of NaOH = 60 mL

mol of HClO4 = Molarity of HClO4 * Volume of HClO4

mol of HClO4 = 0.2 M * 30 mL = 6 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.15 M * 60 mL = 9 mmol

We have:

mol(HClO4) = 6 mmol

mol(NaOH) = 9 mmol

6 mmol of both will react

remaining mol of NaOH = 3 mmol

Total volume = 90.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 3 mmol/90.0 mL

= 0.0333 M

we have below equation to be used:

pOH = -log [OH-]

= -log (3.333*10^-2)

= 1.48

we have below equation to be used:

PH = 14 - pOH

= 14 - 1.48

= 12.52

Answer: 12.52