Question

Consider the titration of 51.0 mL of 0.200 MHNO3 with 0.435 M NaOH.

Part A

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

Express your answer using three significant figures.

Part B

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

Express your answer using three significant figures.

Part C

What is the pH at the equivalence point?

Express your answer using two decimal places.

Answer 1

part A)

Initial mol of HNO3=volume*molarity=51.0 ml*(0.2 mol/L)=0.051L*(0.2mol/L)=0.0102 mol=1.02*10^-2 mol=1.02*10^-2 mol *(1000mmol/mol)=1.02*10^1=10.2 mmol

part B)NaOH +HNO3 ---->NaNO3 +H2O

NaOH and HNO3 react in 1:1 molar ratio

to completely neutralize HNO3,

mol of NaOH required =mol of HNO3 present=0.0102 mol (from part a)

So,volume of NaOH required=mol/molarity=0.0102 mol/0.435 mol/L=0.0234 L=0.0234L*(1000ml/L)=23.4 ml

part C) mol of HNO3 remaining=0

mol of NaOH=0

As complete neutralization of acid takes place,so only NaNO3 salt remaining in the solution ,which is neutral so undergo complete ionization.

pH of neutral solution=7.00