Salt from Weak Acid with Strong Base
Weak acids with weak bases can form total (perfect) hydrolyzed salts in water. Both cations and anions can be hydrolyzed in water. This salt solution can be acidic, basic, or neutral. This depends on the ratio of the strength of cations to anions in reaction with water.
Example
A weak HCN acid is mixed with a weak base, NH 3 will form NH 4 CN salt. HCN is partially ionized in water to form H + and CN - whereas NH 3 in water is partially ionized to form NH4 + and OH-. CN - base anions and NH 4 + acid cations can be hydrolyzed in water.
NH 4 CN (aq) → NH 4 + (aq) + CN - (aq)
NH 4 + (aq) + H 2 O → NH 3 (aq) + H 3 O (aq) +
CN - (aq) + H 2 O (e) → HCN (aq) + OH - (aq)
Salt from Weak Acid with Weak Bases
The nature of the solution depends on the relative strength of acids and their constituent bases (Ka and Kb)
If Ka <Kb (acid is weaker than base) the anion will be hydrolyzed more and the solution is basic.
if Ka> Kb (the acid is stronger than the base) the cation will be hydrolyzed more in an acidic solution.
If Ka = Kb (acid is as weak as base) the solution is neutral.
Examples of Salt Solution
Everyday life
salt and its use
Example Problem 1
The following are some examples along with solving problems related to salt hydrolysis that we have just studied together:
1. What is the pH of the solution of 100 mL 0.01 M sodium cyanide solution? (Head of HCN = 10-10)
Solution and Answers:
Sodium cyanide solution is formed from a mixture of strong bases (NaOH) with weak acids (HCN). Thus, the salt solution undergoes partial hydrolysis and is basic.
NaCN (aq) → Na + (aq) + CN– (aq)
Hydrolyzed ions are CN- ions. The concentration of CN ions is 0.01 M. Thus, the pH of the salt solution can be obtained through the following equation:
[OH-] = {(Kw / Ka) ([hydrolyzed ion])} 1/2
[OH-] = {(10-14 / 10-10) (0.01)} 1/2
[OH-] = 10-3 B
Thus, the pOH of the solution is 3. So, the pH of the salt solution is 11.
2. What is the pH of the solution of 200 mL of 0.1 M barium acetate solution? (Head CH3COOH = 2.10-5)
Solution and Answers:
Barium acetate solution is formed from a mixture of strong bases (Ba (OH) 2) with weak acids (CH3COOH). Thus, the salt solution undergoes partial hydrolysis and is basic.
BA (CH3COO) 2 (aq) → Ba + 2 (aq) + 2 CH3COO– (aq)
Hydrolyzed ions are CH3COO- ions. The CH3COO- ion concentration is 0.2 M. Thus, the pH of the salt solution can be obtained through the following equation:
[OH-] = {(Kw / Ka) ([hydrolyzed ion])} 1/2
[OH-] = {(10-14 / 2.10-5) (0.2)} 1/2
[OH-] = 10-5 billion
Thus, the pOH of the solution is 5. So, the pH of the salt solution is 9.
3. Calculate the pH of the NH4Cl 0.42 M solution! (NH4OH KB = 1.8.10-5)
Solution and Answers:
Ammonium chloride solution is formed from a mixture of weak bases (NH4OH) with strong acids (HCl). Thus, the salt solution undergoes partial hydrolysis and is acidic.
NH4Cl (aq) → NH4 + (aq) + Cl– (aq)
Hydrolyzed ions are NH4 + ions. The concentration of the NH4 + ion is 0.42 M. Thus, the pH of the salt solution can be obtained through the following equation:
[H +] = {(Kw / Kb) ([hydrolyzed ion])} 1/2
[H +] = {(10-14 / 1.8.10-5) (0.42)} 1/2
[H +] = 1,53.10-5 M
Thus, the pH of the salt solution is 4.82.
Example Problem 2
a.First try
Weigh 5.58 grams of table salt (NaCl)
Put salt in a beaker containing ml water, then stir
Put the kitchen salt solution into the measuring flask
Add water until the volume reaches 100 ml, then stir
b. Second try
Put the 3M HCl solution + water into the beaker
Then mix the HCL solution with 100 ml of 1M HCL
After that measure until the volume reaches 100 ml, then stir
Answers a & b
Conclusion
a. First conclusion
From observations and laboratory practices, our group was able to find out that, making a 100 ml NaCl 1M salt solution by dissolving 5.85 grams of salt + water to 100 ml volume. While from the results of this chemical practice, our group obtained concentration data from NaCl solution of 1M
b. Second conclusion
From laboratory observations and practices, our group was able to find out that, making a 100 ml 1M HCL solution by diluting a 33.33 ml HCL 3M + water solution to reach a 100 ml volume.
Suggestion
In practicing the making of salt solutions must be done seriously and thoroughly. Because, if the practicum is not careful or incorrect in calculating the mass, it will affect the process of making the solution. Therefore, in this practicum, it must be careful and thorough.
Example Problem 3
One way to obtain salt compounds is by reacting acids with alkaline substances. This reaction is known as a salting reaction or also called a neutralization reaction. In everyday life salt that is often used include: table salt (NaCl), English salt (MgSO4) as a laxative, baking soda (NaHCO3) as a bread developer, monosodium glutamate (MSG) as a flavor enhancer.
The nature of salt depends on the acid and base forming it. Salt that comes from the reaction between acid and base can be acidic, basic or neutral.
Salt that is acidic, has a pH <7, comes from the reaction between strong acids and weak bases. Example: NH4Cl (ammonium chloride / salmoniac), and NH4NO3 (ammonium nitrate).
Salt that is basic, has a pH> 7, comes from the reaction between a weak acid and a strong base. Examples: KNO2 (potassium nitrite), NaHCO3 (sodium bicarbonate / baking soda), NaCH3COO (sodium acetate), KCN (potassium cyanide / potassium), and KF (potassium phosphate).
Salt that is neutral, has a pH = 7, comes from strong acids and strong bases.
Example: NaCl (sodium chloride), KI (potassium iodide), and KNO3 (potassium nitrate).
salt, NaCl → Na + + Cl-
iron sulfate, Fe2 (SO4) 3 → 2Fe3 + + 3SO3-4