 Lab 11A:  Titration of Hydrochloric Acid                                                                   SCH 3U

# Procedure

## Discussion

Discussion:

Questions –

1.

2.

7.

1.         Determine the number of moles of sodium hydroxide used in each titration.

### Trial #1

nNaOH = 20.7 mL X 0.1 mol/L

= 2.07 mmol

= 0.00207 mol

Trial #2

nNaOH = 20.5 mL X 0.1 mol/L

= 2.05 mmol

= 0.00205 mol

The number of moles of sodium hydroxide used in each titration is 0.00207 mol and 0.00205 mol.

2.         Determine the number of moles of hydrochloric acid used in each titration.

Mole ratio HCl : NaOH = 1 : 1

Trial #1

nHCl = 0.00207 mol X 1/1

= 0.00207 mol

Trial #2

nHCl = 0.00205 mol X 1/1

= 0.00205 mol

The number of moles of hydrochloric acid used in each titration is 0.00207 mol and 0.00205 mol.

3.         Determine the concentration of the hydrochloric acid solution obtained from each titration.  Then average the results.

Trial #1

CHCl = 2.07 mmol/20 mL

= 0.1035 mol/L

Trial #2

CHCl = 2.05 mmol/20 mL

= 0.1025 mol/L

Average = (0.1035 mol/L + 0.1025 mol/L)/2

= 0.103 mol/L

The average concentration of the hydrochloric acid is 0.103 mol/L.

4.         If 27.31 mL of 0.2115 mol/L sodium hydroxide is able to neutralize 37.45 mL of hydrochloric acid, what is the concentration of the acid?

nNaOH = 27.31 mL X 0.2115 mol/L

= 5.78 mmol

nHCl = 5.78 mmol X 1/1

= 5.78 mmol

CHCl = 5.78 mmol/37.45 mL

= 0.154 mol/L

The concentration of the hydrochloric acid is 0.154 mol/L.

5.         What volume of 0.1117 mol/L hydrochloric acid is needed to neutralize 28.67 mL of 0.137 mol/L of potassium hydroxide?

nNaOH = 28.67 mL X 0.137 mol/L

= 3.928 mmol

nHCl = 3.928 mmol X 1/1

= 3.928 mmol

vHCl = 3.928 mmol / 0.1117 mol/L

= 35.17 mL

The volume of hydrochloric acid is 35.17 mL

6.         Why does the pink colour, which forms at the point where the sodium hydroxide comes into contact with the solution in the flask, disappear more slowly near the endpoint?

The pink colour disappears more slowly near the endpoint because the titrant and the sample are almost completely reacted.

7.         Why is it a good idea to carry out titrations in duplicate?

It is a good idea to carry out titrations in duplicate to improve the reliability of the answer.

8.         Why might you be told to use distilled water to wash a drop of solution adhering to one of the buret tips into the Erlenmeyer flask?

Distilled water can be used to wash a drop of solution from the buret tip into the flask because it will not affect the titration.  Since distilled water is an amphiprotic, it is capable of acting as an acid or a base in different chemical reactions.  When distilled water is used to wash a drop of solution adhering to one of the buret tips, water can acts as an acid or base in order to neutralize the drop of solution.

9.         Would the addition of several millilitres of distilled water to the Erlenmeyer flask during a titration affect the results of the titration?  Explain your answer.

Yes, the addition of several milliliters of distilled water will affect the results of the titration.  The addition of distilled water (solvent) will dilute the solution in the Erlenmeyer flask and will affect its concentration

10.       If 35.93 mL of 0.1590 mol/L sodium hydroxide neutralizes 27.48 mL of sulphuric acid, what is the concentration of the sulphuric acid?

H2SO4  +  2NaOH    Na2SO4  +  2H2O

nNaOH = 35.93 mL X 0.1590 mol/L

= 5.71 mmol

Mole ratio H2SO4 : NaOH = 1 : 2

nH2SO4 = 5.71 mmol X ½

= 2.855 mmol

CH2SO4 = 2.855 mmol/27.48 mL

= 0.104 mol/L

The concentration of the sulphuric acid is 0.104 mol/L