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Reactions of Acids and Bases
A neutralization reaction is a double displacement reaction between an acid and a base to produce a salt and water. The salt is formed from the negative ion of the acid and the positive ion of the base. When just enough base (or acid) is added to react exactly with the acid (or base) in a solution, we say the acid has been neutralized. Neutralization reactions occur when neither the acid nor the base are in excess.
The effects of the acid and base are destroyed by the neutralization. The salt produced will determine the pH of the resulting solution.
Neutralization of Strong Acids and Strong Bases
For strong acid and strong base reactions, the salt produced will always be a neutral salt. We will discuss what neutral (and acidic and basic) salts mean more when we introduce weak acids and bases.
Since we are reacting a strong acid with a strong base, you get complete dissociation of the acid and base. So, as in example above, you get the following complete ionic equation:
H+^ + Cl–^ + Na+^ + OH–^ → Na+^ + Cl–^ + H 2 O
If you reduce the reaction by removing spectator ions, the net ionic equation is
H+^ + OH–^ → H 2 O
This will ALWAYS be the net ionic equation for a strong acid – strong base reaction, and the pH of the resulting solutions will ALWAYS be 7.
Another example of such a reaction is HNO 3 and Ca(OH) 2. This type of reaction is important because the base will dissociate twice the number of moles of hydroxide ions. You need to be aware of this. The neutralization reaction will look like this
2HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + 2H 2 O
The total ionic equation becomes
2H+^ + 2NO 3 –^ + Ca+^ + 2OH–^ → Ca+^ + 2NO 3 –^ + 2H 2 O
If you reduce the reaction by removing spectator ions, the net ionic equation is
2H+^ + 2OH–^ → 2H 2 O
And if you simplify both sides of the reaction, your net ionic equation is
H+^ + OH–^ → H 2 O
This will ALWAYS be the net ionic equation for a strong acid – strong base reaction, and the pH of the resulting solutions will ALWAYS be 7.
Sulfuric acid is an oddball…
The only exception to the ideas of strong acids and bases and neutralization reactions has to do with sulfuric acid. We have already discussed that only the first hydrogen in sulfuric acid completely dissociates. You may see problems, or examples in textbooks (or online) were they could give you the reaction shown below.
You will ONLY see things like this if the problem or situation is neutralization with the same number of moles of H+^ and OH–. However, the pH of this solution at equivalence will NOT be 7, and the net ionic equation will not be the formation of water. Again, this will come later.
Neutralization of Weak Acids and Bases
For this, let’s look at the example of HCl being neutralized with Al(OH) 3. The neutralization reaction will be
3HCl + Al(OH) 3 → AlCl 3 + 3H 2 O
This reaction has a strong acid with a weak base. You would still do the calculations for the neutralization the same as if it were a strong acid – strong base problem. Because of the nature of this being strong acid and weak base, the final solution when neutralization occurs will be acidic and the pH will be less than 7.
In these types of problems, you are relying on the ideal that the number of moles of acid (H+) is equal to the number of moles of base (OH–). You will be attempting to solve for an unknown molarity or volume.
A laboratory procedure used to determine the unknown molarity of a solution is called a titration.
Example. What volume of 0.0947 M NaOH is needed to neutralize 21.4 mL of 0.106 M HCl?
HCl + NaOH → NaCl + H 2 O molA = molB MAVA = MBVB (0.106 M)(21.4 mL) = (0.0947 M)VB VB = 24.0 mL
Acid and Base Reactions 1
Neutralization.
- What concentration of HCl will be required if you use 100.0 mL of that acid to neutralize 0.010L of 3.0M sodium hydroxide?
- What volume of 0.50M Ba(OH) 2 is required to be mixed with 35.0 mL of 0.60M HCl to have a resulting solution with a pH = 7?
- What volume of 5.0M H 2 SO 4 is required to neutralize a NaOH solution in which 2.50 g of NaOH (MM = 40 g/mol) has been dissolved in water to produce 100.0 mL of NaOH solution?
- Determine the molarity of a monoprotic acid solution in which 40.0 mL of the acid has been neutralized by 120.0 mL of 0.531M NaOH.
- 0.275 grams of solid NaOH (MM = 40 g/mol) was added to 35.4 mL of HCl. Complete neutralization occurred. What was the molarity of the HCl?
- What would be the volume of 0.250M monoprotic acid required to react with 0.500 grams of Ca(OH) 2?
- What is the molarity of an oxalic acid (H 2 C 2 O 4 ) solution if 22.50 mL of this solution requires 35. mL of 0.198 M NaOH for complete neutralization?
- A solution was prepared by dissolving 25.0 grams of Ba(OH) 2 in water to make one liter of solution. How many milliliters of 0.200M H 2 SO 4 would be required to react with 25.0 mL of the Ba(OH) 2 solution?
- What is the molarity of Ca(OH) 2 if 200.00 mL of Ca(OH) 2 is required to completely neutralize 100.00 mL of 0.350M H 3 PO 4?
Not Neutralization
- You begin with 25 mL of 0.2 M HCl. You titrate this solution with 0.1 M NaOH. Calculate the pH when you have added 0 mL, 20 mL, 40 mL, 50 mL, 60 mL and 80 mL of NaOH.
