Determine the pH of a 0.62M NH4NO3 solution at 25oC. The Kb for NH3 is 1.76x10-5. Calculate Ka
pH = 4.73
To determine the pH of a NH4NO3 solution, you need to consider the dissociation of NH4NO3 into NH4+ and NO3- ions, as well as the hydrolysis of NH4+ ion.
Step 1: Write the dissociation equation for NH4NO3:
NH4NO3 ⇌ NH4+ + NO3-
Step 2: Calculate the concentration of NH4+ and NO3- ions:
Since NH4NO3 is a strong electrolyte, it fully dissociates to give 0.62M NH4+ and 0.62M NO3- ions.
Step 3: Determine the hydrolysis reaction of NH4+:
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Step 4: Calculate the concentration of NH3 produced from the hydrolysis of NH4+:
Assuming the concentration of NH4+ at equilibrium is x, the concentration of NH3 and H3O+ ions will be x as well.
Step 5: Write the expression for the equilibrium constant for the hydrolysis reaction:
Kb = [NH3][H3O+]/[NH4+]
Given Kb = 1.76 x 10^-5
Step 6: Substitute the known values into the equilibrium constant expression:
1.76 x 10^-5 = (x)(x)/(0.62-x)
Step 7: Solve for x:
Since x is small compared to 0.62, it can be assumed that 0.62 - x ≈ 0.62.
1.76 x 10^-5 = x^2 / 0.62
Rearranging the equation:
0.62 * 1.76 x 10^-5 = x^2
Simplifying:
1.09 x 10^-5 = x^2
Taking the square root:
x = 1.04 x 10^-3
Step 8: Calculate the concentration of NH3:
Since the concentration of NH3 produced is x = 1.04 x 10^-3, the concentration of NH3 is also 1.04 x 10^-3 M.
Step 9: Calculate the concentration of H3O+ ions:
Since the concentration of H3O+ ions is the same as NH3, it is also 1.04 x 10^-3 M.
Step 10: Calculate the pH of the solution:
pH = -log[H3O+]
pH = -log(1.04 x 10^-3)
pH = 2.98
Therefore, the pH of a 0.62 M NH4NO3 solution at 25oC is approximately 2.98.
To calculate the Ka, please provide the necessary information, as the given data does not directly allow for the calculation of Ka.
To calculate the pH of a solution, we need to know the concentration of the acid or base and the dissociation constant (Ka or Kb) of the relevant species.
In this case, we are dealing with NH4NO3, which is an ionic compound that dissociates into NH4+ (ammonium) and NO3- (nitrate) ions in water. Since NH4+ is the conjugate acid of NH3, we can use the Kb value for NH3 to find the Ka value for NH4+.
Let's start by finding the concentration of NH4+ in the solution. Since NH4NO3 is a strong electrolyte, it will dissociate completely, so the concentration of NH4+ is the same as the concentration of NH4NO3.
Given:
Concentration of NH4NO3 = 0.62 M
Next, we need to find the value of Ka. Ka is related to Kb through the autoionization constant of water (Kw), which is 1.0 x 10^-14 at 25°C:
Kw = Ka * Kb
Solving for Ka:
Ka = Kw / Kb
Substituting the values:
Ka = (1.0 x 10^-14) / (1.76 x 10^-5)
Calculating the Ka value gives:
Ka = 5.68 x 10^-10
Now we have the concentration of NH4+ (0.62 M) and the Ka value (5.68 x 10^-10) for NH4+. We can use this information to calculate the pH of the solution.
The dissociation of NH4+ can be represented as follows:
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
Since NH4+ is a weak acid, we can assume that the concentration of NH4+ that dissociates is negligible compared to the initial concentration. So, we can say that the concentration of NH4+ that dissociates is "x" and the concentration of NH4+ that remains is (0.62 - x).
Using the Ka expression for NH4+:
Ka = [NH3][H3O+] / [NH4+]
Substituting the values:
5.68 x 10^-10 = x * x / (0.62 - x)
Since x is assumed to be much smaller compared to the initial concentration (0.62), we can approximate (0.62 - x) as 0.62:
5.68 x 10^-10 = x^2 / 0.62
Rearranging this equation:
x^2 = 5.68 x 10^-10 * 0.62
Solving for x:
x = sqrt(5.68 x 10^-10 * 0.62)
Calculating the value gives:
x ≈ 2.51 x 10^-6
Since x represents the concentration of H3O+ ions, we can use this value to calculate the pH of the solution:
pH = -log[H3O+]
pH = -log(2.51 x 10^-6)
Calculating the pH value gives:
pH ≈ 5.60
Therefore, the pH of the 0.62M NH4NO3 solution at 25°C is approximately 5.60.