Subjects
| Sections | |||
|---|---|---|---|
| Buffer | 50 mins | 0 completed | Learn |
| Acid and Base Titration Curves | 32 mins | 0 completed | Learn |
| Weak Acid Strong Base Titrations | 34 mins | 0 completed | Learn |
| Weak Base Strong Acid Titrations | 17 mins | 0 completed | Learn |
| Strong Acid Strong Base Titrations | 10 mins | 0 completed | Learn Summary |
| Titrations of Diprotic and Polyprotic Acids | 27 mins | 0 completed | Learn |
| Ksp | 58 mins | 0 completed | Learn |
| Additional Practice |
|---|
| Equivalence Point |
| Acid Base Indicators |
| Selective Precipitation |
| Formation Constant |
| Additional Guides |
|---|
| Henderson Hasselbalch Equation |
A chemistry student weighs out 0.0899 g of acetic acid, HC 2H3O2 into a 250 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1900 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point.
a) 3.07 mL
b) 4.08 mL
c) 5.48 mL
d) 6.15 mL
e) 7.88 mL
We’re being asked to calculate the volume of 0.1900 M NaOH required to titrate 0.0899 g acetic acid (HC2H3O2) to the equivalence point.
The reaction between HC2H3O2 and NaOH will be a double displacement reaction where H+ reacts with OH– and C2H3O2– reacts with Na+.
The overall reaction is:
HC2H3O2 + NaOH → H2O + NaC2H3O2
Recall that at the equivalence point of a titration:
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