Examples of Salt Hydrolysis Problems
1. Concentration of NH4Cl 0.4 M, if kb NH3 = 10-5
2. The pH of NH4Cl 0.4 M is (kb NH3 = 10-5)
The following are some examples along with solving problems related to salt hydrolysis that we have just studied together:
What is the pH of the solution from 100 mL 0.01 M sodium cyanide solution? (Head of HCN = 10-10)
Settlement:
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 M
Thus, the pOH of the solution is 3. So, the pH of the salt solution is 11.
What is the pH of the solution from 200 mL of 0.1 M barium acetate solution? (Head CH3COOH = 2.10-5)
Settlement:
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 M
Thus, the pOH of the solution is 5. So, the pH of the salt solution is 9.
Calculate the pH of the NH4Cl 0.42 M solution! (NH4OH KB = 1.8.10-5)
Settlement:
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.
Calculate the pH of the 2.00 M NH4CN solution! (Head of HCN = 4,9.10-10 and Kb NH4OH = 1,8.10-5)
Settlement:
Ammonium cyanide solution is formed from a mixture of a weak base (NH4OH) with a weak acid (HCN). Thus, the salt solution undergoes total hydrolysis.
NH4Cl (aq) → NH4 + (aq) + CN– (aq)
Hydrolyzed ions are NH4 + ions and CN- ions. Thus, the pH of the salt solution can be obtained through the following equation:
[H +] = {Kw (Ka / Kb)} 1/2
[H +] = {10-14 (4,9.10-10 / 1,8.10-5)} 1/2
[H +] = 5,22.10-10 M
Therefore, the pH of the salt solution is 9.28.
What mass of NaCN salt must be dissolved to form 250 mL of solution with a pH of 10? (Head of HCN = 10-10 and Mr. NaCN = 49)
Settlement:
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)
pH = 10, means pOH = 4
Thus, [OH–] = 10-4 AD
The calculation of the pH of the saline solution can be obtained through the following equation:
[OH–] = {(Kw / Ka) ([hydrolyzed ion])} 1/2
10-4 = {(10-14 / 10-10) [hydrolyzed ion]} 1/2
[hydrolyzed ion] = 10-4 M
The required NaCN salt concentration is 10-4 M. The volume of the solution is 250 mL = 0.25 L. Thus, the mole of NaCN salt needed is:
Mol = Volume x Molar
Mol = 0.25 x 10-4 = 2.5 x 10-5 mol
So, the mass of NaCN salt needed is 2.5 x 10-5 x 49 = 1,225 x 10-3 grams = 1,225 mg.
That is the full review Hopefully what is discussed above is useful for readers. That is all and thank you.