Question

29).

A 50.00-mL sample of 0.100 M KOH is titrated with 0.100 M HNO₃. Calculate the pH of the solution after 52.00 mL of HNO₃ is added.

1.29

2.71

11.29

12.71

None of these choices are correct.

Answer 1

Number of moles of KOH , n = Molarity x volume in L

                                          = 0.100 M x 0.050 L

                                         = 0.005 moles

Number of moles of HNO₃ added , n' = Molarity x volume in L

                                                      = 0.100 Mx 0.052 L

                                                      = 0.0052 moles

The balanced reaction is KOH + HNO₃ KNO₃ + H₂O

According to the balanced equation '

1 mole of KOH reacts with 1 moles of HNO₃

0.005 mole of KOH reacts with 0.005 moles of HNO₃

So 0.0052 - 0.005 = 0.0002 moles of HNO₃ left unreacted in the solution which gives acidic nature to the resulting solution.Also 1 mole of HNO₃ contains 1 mole of H⁺ ---> [H⁺]_mol = [HNO₃]_mol = 0.0002 moles

Total volume of the solution , V = 50.0+52.0 = 102.0 mL = 0.102 L

So concentration of H⁺ in the resultant solution = number of moles of H⁺ left / total volume in L

                                                                      = 0.0002 mol / 0.102 L

                                                                      = 1.96x10^-3 M

pH = - logH⁺

     = - log (1.96x10^-3 )

     = 2.71

Therefore the pH of the resulting solution is 2.71