| 1. |
What strong acid and strong base
combine in a neutralization reaction to yield Li2SO4? |
| |
| Strong
acid: H2SO4 |
Strong
base: LiOH2 |
|
|
| 2. |
Pick an acid and a base from the list below.
Write a balanced chemical equation for the neutralization
reaction using the acid and base you chose. (I know you can't
enter subscripts in the box.) |
| |
|
| |
(Any
combination of acid and base chosen from the list above
would balance the same way because all the acids are monoprotic
and all the bases are dibasic.)
2 Acid + Base --> Salt + 2 water
Example: 2 HNO3
+ Mg(OH)2--> Mg(NO3)2
+ 2 H2O |
| 3. |
Assume that your base is available as an
aqueous solution with a concentration of 0.200 M. What volume
of your base will it take to neutralize 25.0 mL of your acid
if your acid is 0.300 M. (Note: The answer is not
37.5 mL) You must explain the steps you used in your calculation. |
| |
Facts from the question:
acid: 25.0 mL of 0.300 M
HNO3, 0.300 N HNO3
base: 0.200 M Mg(OH)2,
or 0.400 N Mg(OH)2
NAVA
= NBVB (Where N is normality and V
is volume)
(0.300 N HNO3)(25.0
mL)=(0.400 N Mg(OH)2)(volume
of base)
volume of base = 18.8 mL
(Any
combination of acid and base chosen from the list above
would give the same result because all the acids are monoprotic
and all the bases are dibasic.)
|
| 4. |
At the end of your titration in question
3, what substances are present? (Hint: Is there any acid remaining?
Any base?) |
| |
Only your salt and water
remain. |
| |
|
| 5. |
Consider a titration of 30 mL of acetic acid
(0.200 M) with sodium hydroxide (0.200 M). What volume of
NaOH is needed to neutralize the acid? |
| |
CH3COOH
+ NaOH --> CH3COONa
+ H2O
Facts from the question:
acid: 30.0 mL of 0.200 M CH3COOH
or 0.200 N CH3COOH
base: 0.200 M NaOH
or 0.200 N NaOH
NAVA
= NBVB (Where N is normality and V
is volume)
(0.200
N CH3COOH)(30.0
mL)=(0.200
N NaOH)(volume
of base)
volume of base = 30.0 mL |
| 6. |
What is the salt you are forming? What is
the conjugate acid of the anion of this salt? |
| |
CH3COOH
+ NaOH --> CH3COONa
+ H2O
The salt is sodium acetate
(CH3COONa).
The anion is the acetate ion (CH3COO–).
The conjugate acid of the acetate ion is acetic acid, (CH3COOH). |
| 7. |
What volume of sodium hydroxide would be
used to get half-way to neutralization? |
| |
15.0 mL
(It would take 30.0 mL to be neutralized, see the calculation
in question 5. Half of 30.0 mL is 15.0 mL.) |
| 8. |
What substances are present in your flask
when you have neutralized half of the acid? |
| |
Acetic acid (CH3COOH)
and the acetate ion (CH3COO–) |
| 9. |
What is the name for the mixture in question
8? (Hint: See page 280 in your text.) |
| |
A buffer. (A weak acid and
its conjugate base form a buffer.) |
