Question

If 10.0 ml of 0.100 M strong base is titrated with 25.0 ml of 0.100 M strong acid, what is the pH of the resulting solution?

Answer 1

Number of mmol of strong base , n = Molarity x volume in mL

                                                   = 0.100 M x 10.0 mL

                                                   = 1.0 mol

Number of mmol of strong acid , n = Molarity x volume in mL

                                                   = 0.100 M x 25.0 mL

                                                   = 2.5 mol

Since both are strong 1 equivalent mol of acid reacts with 1 equivalent mol of a base

So 1.0 mole of acid reacts completly with 1.0 mol of base.

2.5 - 1.0 = 1.5 mole of acid will left unreacted in the mixture solution.

So the concentration of acid left unreacted ,

So [H⁺]= 0.043 M

pH = - log [H⁺]

     = -log 0.043

     = 1.37

The pH of the resulting solution is 1.37