Question

What is the pH of a solution in which 52 mL of 0.20M NaOH is added to 16 mL of 0.160M HCl?

Answer 1

Solution :-

Lets first calculate the moles of NaOH and HCl using the molarity and volume

Moles of NaOH = 0.20 mol per L * 0.052 L

                           = 0.0104 mol

Moles of HCl = 0.160 mol per L * 0.0160 L

                       = 0.00256 mol HCl

Moles of HCl are less than moles of NaOH so the HCl is limiting reactant

Now lets find the moles of the NaOH remain after the reaction

Moles of NaOH remain = 0.0104 mol – 0.00256 mol = 0.00784 mol NaOH

Total volume = 52 ml + 16 ml = 68 ml = 0.068 L

New molarity of the NaOH = moles / volume in liter

                                              = 0.00784 mol /0.068 L

                                              = 0.1153 M

Now lets calculate the pOH

pOH=-log [OH-]

        = -log [0.1153]

        = 0.938

pH+ pOH = 14

pH= 14 – pOH

     = 14 – 0.938

    = 13.1

So the pH is 13.1