Question

Consider the titration of 50.0 mL of 0.130 MHNO3 with 0.450 M NaOH.

How many millimoles of HNO3 are present at the start of the titration?

How many milliliters of NaOH are required to reach the equivalence point?

What is the pH at the equivalence point?

Answer 1

1)Consider the titration of 50.0 mL of 0.130 MHNO3 with 0.450 M NaOH.

How many millimoles of HNO3 are present at the start of the titration?

Solution -:-

Volume of HNO3 = 50.0 ml = 0.050 L

Molarity of HNO3 = 0.130 M

Lets calculate the moles of the HNO3 using the molarity and volume

Moles = molarity * volume in liter

Moles of HNO3 = 0.130 mol per L * 0.050 L = 0.0065 mol HNO3

Now lets convert moles to millimoles

0.0065 mol HNO3 *1000 millimoles / 1 mol = 6.5 millimoles of HNO3

Therefore initially 6.5 millimoles of HNO3 are present.

2)How many milliliters of NaOH are required to reach the equivalence point?

Solution :- balanced reaction equation is as follows

HNO3 +NaOH ------ > NaNO3 + H2O

Since mole ratio of the HNO3 and NaOH is 1 :1 therefore the volume of the NaOH needed to reach the equivalence point is calculated as follows

Volume of NaOH = molarity of HNO3 * volume of HNO3 / molarity of NaOH

                              = 0.130 M * 50.0 ml / 0.450 M

                             = 14.44 ml

14.44 milliliter of NaOH is needed to reach the equivalence point.

3)What is the pH at the equivalence point?

Solution :-

HNO3 and NaOH are strong acid and strong base therefore they neutralizes the each other by forming neutral salt and water

Therefore at the equivalence point all the acid and base react completely therefore solution becomes neutral that is having pH =7