Question

A 25.0 mL sample of 0.100 HClO4 is titrated with a 0.100 M NaOH solution. What is the pH after the addition of 10.0 mL of NaOH? (HClO4 is a strong acid)

Answer 1

Solution :-

We are given with the molarity and volume of the acid and base both are strong acid and strong base and they have 1 :1 mole ratio as follows

HClO4 + NaOH ---- > NaClO4 + H2O

So lets calculate the initial moles of the acid and base

Moles = molarity * volume in liter

Moles of HClO4 = 0.100 mol per L * 0.025 L

                            = 0.0025 mol HClO4

Moles of NaOH =0.100 mol per L * 0.010 L

                           = 0.001 mol NaOH

Moles of NaOH are less than moles of HClO4 therefore NaOH is the limiting reactant and hence react completely

Since HClO4 is excess reactant so lets calculate moles of HClO4 remaining after the reaction

Moles of HClO4 remaining = 0.0025 mol – 0.001 mol = 0.0015 mol HClO4

Now lets calculate new molarity of the HClO4 at total volume

Total volume = 25 ml + 10 ml = 35 ml = 0.035 L

New molarity of the HClO4 = 0.0015 mol / 0.035 L = 0.0429 M

Since HClO4 is strong acid therefore it dissociate completely to give 0.0429 M H+

So lets calculate the pH using this concentration

pH= -log[H+]

pH =-log [0.0429]

pH = 1.37

There for pH after titration with 10 ml 0.100 M NaOH is 1.37