SALT HYDROLYSIS
1.) Calculate the pH of a 0.24M sodium formate (HCOONa) solution. Kb=5.9x10^-11.
2.) Calculate the pH of 0.25M pyridium chloride (C5H5NHCl)solution. Ka=5.9x10^6.
1.) The pH of a 0.24M sodium formate solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([salt]/[acid])
pKa = -log(Kb) = -log(5.9x10^-11) = 10.22
pH = 10.22 + log(0.24/0.24) = 10.22
2.) The pH of a 0.25M pyridium chloride solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([salt]/[acid])
pKa = -log(Ka) = -log(5.9x10^6) = 6.23
pH = 6.23 + log(0.25/0.25) = 6.23
To calculate the pH of a solution involving salt hydrolysis, we need to consider the hydrolysis reaction of the salt and the equilibrium constants (Kw, Ka or Kb) associated with it.
1.) Calculation of the pH of a 0.24M sodium formate (HCOONa) solution:
The hydrolysis reaction for sodium formate is as follows:
HCOONa + H2O ⇌ HCOOH + NaOH
Step 1: Write the chemical equation for the hydrolysis reaction of the salt:
HCOONa + H2O ⇌ HCOOH + NaOH
Step 2: Determine the initial concentration of the sodium formate and the change in concentration for each species as the reaction proceeds:
Initial Concentration:
[HCOONa] = 0.24 M
Change in Concentration:
[HCOONa] = -x M
[HCOOH] = +x M
[NaOH] = +x M
Step 3: Write the expression for the equilibrium constant (Kb) and substitute the equilibrium concentrations into the expression:
Kb = ([HCOOH] * [NaOH]) / [HCOONa]
Kb = x * x / (0.24 - x)
Since the concentration of sodium formate (HCOONa) is much greater than the change in concentration (x) due to hydrolysis, it can be approximated as:
Kb = x * x / (0.24)
Step 4: Solve the quadratic equation for x using the given Kb value:
5.9x10^-11 = x * x / (0.24)
Rearrange the equation:
x * x = (5.9x10^-11) * (0.24)
x * x = (1.41x10^-11)
Take the square root of both sides:
x = 3.76x10^-6
Step 5: Calculate the concentration of hydroxide ions ([OH-]) using the value of x:
[OH-] = [NaOH] = x = 3.76x10^-6 M
Step 6: Calculate the concentration of hydronium ions ([H3O+]) using the ion product of water (Kw):
Kw = [H3O+] * [OH-]
1.0x10^-14 = [H3O+] * (3.76x10^-6)
[H3O+] = 1.0x10^-14 / (3.76x10^-6)
[H3O+] = 2.66x10^-9 M
Step 7: Calculate the pH using the formula:
pH = -log[H3O+]
pH = -log(2.66x10^-9)
pH ≈ 8.57
Therefore, the pH of the 0.24M sodium formate (HCOONa) solution is approximately 8.57.
2.) Calculation of the pH of a 0.25M pyridinium chloride (C5H5NHCl) solution:
The hydrolysis reaction for pyridinium chloride is as follows:
C5H5NHCl + H2O ⇌ C5H5NH3+ + Cl-
This hydrolysis reaction can be considered as a reaction of a weak acid (C5H5NH3+) with water, and we can use the Ka value to determine the pH.
Step 1: Write the chemical equation for the hydrolysis reaction of the salt:
C5H5NHCl + H2O ⇌ C5H5NH3+ + Cl-
Step 2: Determine the initial concentration of the pyridinium chloride and the change in concentration for each species as the reaction proceeds:
Initial Concentration:
[C5H5NHCl] = 0.25 M
Change in Concentration:
[C5H5NHCl] = -x M
[C5H5NH3+] = +x M
Step 3: Write the expression for the equilibrium constant (Ka) and substitute the equilibrium concentrations into the expression:
Ka = ([C5H5NH3+] / [C5H5NHCl]) = x / (0.25 - x)
Step 4: Solve the quadratic equation for x using the given Ka value:
5.9x10^6 = x / (0.25 - x)
Rearrange the equation:
x = (5.9x10^6) * (0.25 - x)
x = 1.475x10^6 - 5.9x10^6x
6.9x10^6x = 1.475x10^6
x ≈ 0.2133 M
Step 5: Calculate the concentration of hydronium ions ([H3O+]) using the concentration of the acid (C5H5NH3+):
[H3O+] = [C5H5NH3+] = 0.2133 M
Step 6: Calculate the pH using the formula:
pH = -log[H3O+]
pH = -log(0.2133)
pH ≈ 0.67
Therefore, the pH of the 0.25M pyridinium chloride (C5H5NHCl) solution is approximately 0.67.
To calculate the pH of a solution involving salt hydrolysis, we need to consider the reaction of the salt with water and the equilibrium constant associated with that reaction. In both cases, we will need to use the equilibrium constant expression and solve for the concentration of hydroxide ions (OH-) or hydronium ions (H3O+), and then convert it to pH.
1.) Sodium Formate Solution (HCOONa):
The balanced equation for the hydrolysis reaction of sodium formate (HCOONa) in water is given by:
HCOONa + H2O ⇌ HCOOH + NaOH
First, let's write the equilibrium constant expression, which is the ratio of product concentrations to reactant concentrations:
Kb = [HCOOH][NaOH] / [HCOONa]
Since the concentration of the salt (HCOONa) is provided as 0.24M, and we assume complete dissociation,
[HCOONa] = 0.24M
So, the equilibrium constant expression becomes:
Kb = [HCOOH][NaOH] / 0.24
Now, we need to make the assumption that the reaction has reached equilibrium, and the concentration of [HCOOH] and [NaOH] are the same. Let's call this concentration 'x'.
Kb = (x)(x) / 0.24
The Kb value given is 5.9x10^-11, so we can set up the equation to solve for 'x':
5.9x10^-11 = (x)(x) / 0.24
Rearranging the equation and solving for 'x':
(x)(x) = 5.9x10^-11 * 0.24
x^2 = 1.416x10^-11
x ≈ 1.19x10^-6 (concentration of [HCOOH] and [NaOH])
The concentration of hydroxide ions (OH-) is equal to the concentration of NaOH since they have a 1:1 stoichiometric ratio.
[OH-] = 1.19x10^-6 M
To find the concentration of hydronium ions (H3O+), we need to apply the water autoionization constant (Kw):
Kw = [H3O+][OH-]
1.0x10^-14 = [H3O+][1.19x10^-6]
Solving for [H3O+]:
[H3O+] = 1.0x10^-14 / 1.19x10^-6 ≈ 8.4x10^-9 M
Now, we can calculate the pH using the equation:
pH = -log[H3O+]
pH ≈ -log(8.4x10^-9)
pH ≈ 8.08
Therefore, the pH of a 0.24M sodium formate (HCOONa) solution is approximately 8.08.
2.) Pyridium Chloride Solution (C5H5NHCl):
Similar to the previous example, we will write the balanced equation for the hydrolysis reaction:
C5H5NHCl + H2O ⇌ C5H5NH2 + H3O+ + Cl-
The equilibrium constant expression can be written as:
Ka = [C5H5NH2][H3O+] / [C5H5NHCl]
We assume complete dissociation for the salt (C5H5NHCl), so:
[C5H5NHCl] = 0.25M
Substituting this information into the equilibrium constant expression, we get:
Ka = [C5H5NH2][H3O+] / 0.25
Since Ka is given as 5.9x10^6, we set up the equation:
5.9x10^6 = (x)(0.25)
x ≈ 1.5x10^-6 M (concentration of [C5H5NH2])
The concentration of hydronium ions (H3O+) is equal to the concentration of [C5H5NH2] since they have a 1:1 stoichiometric ratio.
[H3O+] = 1.5x10^-6 M
Finally, we can calculate the pH using the equation:
pH = -log[H3O+]
pH ≈ -log(1.5x10^-6)
pH ≈ 5.82
Therefore, the pH of a 0.25M pyridium chloride (C5H5NHCl) solution is approximately 5.82.