|
Lab 11A:
Titration of Hydrochloric Acid
SCH 3U |
|
Purpose
Apparatus
Procedure
Observations
Discussion
|
Discussion:
Questions – 1. Determine the number of moles of sodium hydroxide used in
each titration. 2. Determine the number of moles of hydrochloric acid used in
each titration. 7. Why is it a good idea to carry out titrations in duplicate? 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 < Back to Observations | Proceed to Conclusion > |