Chapter 15 Lecture Problems
- For a 0.100 M CH3COOH at equilibrium, calculate Ka if at equlibrium [H3O] = 1.318*10-3 M:
|
CH3COOH |
H2O |
<--> |
CH3COO- |
H3O+ |
Initial (M) |
0.100 |
|
|
0 |
0 |
Change (M) |
-1.318*10-3 |
|
|
+1.318*10-3 |
+1.318*10-3 |
Final (M) |
0.0987 |
|
|
1.318*10-3 |
1.318*10-3 |
- For a nitrous acid (Ka = 4.5 x 10-4) solution find the pH and pOH for initial nitrous acid concentrations of
- 1.0 M
- 0.001 M
- 10-8 M.
- Solutions
- Acetic Acid Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of acetic acid is added to 250.0 mL of water.
- Solution
- Sodium Acetate Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of sodium acetate is added to 250.0 mL of water.
- Solution
- Ammonium Chloride Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of ammonium chloride is added to 250.0 mL of water.
- Solution
- Nitrous Acid Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of nitrous acid is added to 250.0 mL of water.
- Solution
- Sodium Nitrite Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of sodium nitrite is added to 250.0 mL of water.
- Solution
- Nitric Acid Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of nitric acid is added to 250.0 mL of water.
- Solution
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