Title | Worksheet The Titration of Various Acids and Bases |
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Course | General Chemistry 3 |
Institution | Portland State University |
Pages | 3 |
File Size | 174 KB |
File Type | |
Total Downloads | 54 |
Total Views | 159 |
Worksheet and answers...
The Titration of Various Acids and Bases Worksheet Worksheet:: Ana Waxer 04 04/29/2021 /29/2021
Part 1: Strong Acid with Strong Base
Figure 1: Titration Curve of a strong acid with a strong base. The curve represents an idealized titration of an equal amount strong acid (HCL) and strong monoprotic base (NaOH).
Estimated volume of titrant at equivalence point = ____25 25 25_____mL Estimated pH at equivalence point = ____ ____7_____ 7_____
Part 2: Weak Acid with Strong Base
Figure 2: Titration curve of a weak acid with a strong base. The curve represents the idealized titration of 0.1894 M CH3COOH, a weak acid and 0.2006 M NaOH, a strong base. Estimated volume of titrant at equivalence point = _____25 25 25____ ____ ____mL Estimated pH at equivalence point = _____9 9____
Part 3. Strong Acid with Weak Base Titration Curve
Figure 3: Titration curve of a strong acid with a weak base. The curve represents an idealized titration of 0.30 M HCL, a strong acid and 0.40 M NaHCO3, a weak base. Estimated volume of titrant at equivalence point = ___32 32 32______mL Estimated pH at equivalence point = ____3 3_____
Results and Calculati Calculations: ons: Here is a generic reaction that can be used to describe each of the three trials: acid (aq) + base (aq) → H2O (l) + ionic compound (aq) Using the above reaction, the information provided in the procedure for each reaction, and some stoichiometry to calculate the volume of the analyte that was used in each of the three titrations.
Part 1: Strong Acid with Strong Base
Volume of titrant = __50____ mL
0.3M NaOH x 0.050L = 0.3M HCL x V → V= 0.3M x 0.050L 0.30M V= 0.050L = 50mL
Part 2: Weak Acid with Strong Base Volume of titrant = ___50___ mL 0.2006M NaOH x 0.050L = 0.1894m HCl x V → V= 0.2006M x 0.050L 0.1894M V= 0.0529L = 53mL Part 3. Strong Acid with Weak Base
Volume of titrant = __50____ mL
0.3M HCl x 0.050L = 0.4M NaHCO3 x V → V= 0.3M x 0.050L 0.4M V= 0.0375L = 38mL
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