Question

What volume of a 0.305 M nitric acid solution is required to neutralize 18.3 mL of a 0.112 M potassium hydroxide solution?

__________ mL nitric acid

Answer 1

Number of moles of potassium hydroxide(KOH) is , n = Molarity x volume in L

                                                                                     = 0.112 M x (18.3x10^-3 L)

                                                                                     = 2.05 x 10^-3 mol

The reaction between nitric acid(HNO3) & potassium hydroxide(KOH) is

   HNO₃ + KOH KNO₃ + H₂O

This is the balanced equation.

According to this 1 mole of HNO₃ reacts with 1 mole of KOH

                            2.05 x 10^-3 mole of HNO₃ reacts with 2.05 x 10^-3 mole of KOH

So 2.05 x 10^-3 moles of nitric acid required.

Volume of Nitric acid required is , V = number of moles / Molarity of solution

                                                        = 2.05 x 10^-3 mol / 0.305 M

                                                       = 6.7x10 ^-3 L

                                                      = 6.7 mL                        Since 1mL = 10 ^-3 L

So the volume of nitric acid required is 6.7 mL