Question

Two burettes are set up. The first contains 0.15M NaOH and the second contains an HCl solution of unknown concentration. HCl is dispensed into the reaction flask and Phenolphthalein is added as an indicator. At the end of the titration, 22.2ml of NaOH and 9.45ml HCl were used.

Required:
A. Write a balanced equation for the Acid-Base reaction.
B. What is the total volume of NaOH used in the titration?
C. How many moles of NaOH was used?
D. What is the total volume of HCl used in the titration?
E. What is the final concentration of the HCl solution?

Answer 1

A: NaOH + HCl = NaCl +H2O

B: Volume of NaOH = 22.2ml

C:
`mole=3.33*10^(-3)`

D: Volume of HCl = 9.45 ml

E:
`molarity=0.35 mole/litre = final concentration of HCl`

Step-by-step explanation:

Part A: Balanced chemical equaion

NaOH + HCl = NaCl +H2O

part B:

Volume of NaOH used in titration is 22.2 ml because volume is taken upto the end point of the titration.

Part C:

Calculation of moles of NaOH used:

`mole=volume in litre* molarity`

`mole=(22.2)/(1000) *0.15`

`mole=3.33*10^(-3)`

Part D:

Volume of HCl used in titration is 9.45 ml because volume is taken upto the end point of the titration.

Part E:

`N_1V_1=N_2V_2`

`0.15*22.2=N_2* 9.45`

`N_2=0.35`

`Normality =molarity * n-factor`

But in case of acid it is known as basicity and in case of base it is known as acidity.

in case of NaOH and HCl n-factor is 1;

hence

Normality=molarity;

`molarity=0.35 mole/litre = final concentration of HCl`