Calculate the pH of a 0.44 M aqueous solution
of NH4Cl. Kb = 1.8 × 10−5
.
I've tried used the Ka=Kw/Kb equation then used the resulting 5.5e-10 in Kw=([H+][Cl-])/[NH4Cl] , using that to get the [H+], but it isn't correct and I don't know what else to do.
Correction. I meant Ka=([H+][Cl-])/[NH4Cl] not Kw
You're right that Ka = Kw/Kb = 5.55E-10
But your equation is wrong.
Here is the hydrolysis equation which you need.
.........NH4^+ + H2O ==> NH3 + H3O^+
I........0.44.............0......0
C.........-x..............x......x
E.......0.44-x............x.......x
Then Ka = (x)(x)/(0.44-x)
Solve for x = (H3O^+) and convert to pH. Most NH4Cl solution are somewhere around pH 5 so you should get close to that (but probably not 5.0).
To calculate the pH of a solution of NH4Cl, we need to consider the reaction between NH4Cl and water. NH4Cl is a salt that contains the ammonium ion (NH4+) and chloride ion (Cl-). In water, NH4+ acts as a weak acid (donates a proton) and Cl- is a spectator ion.
The reaction between NH4+ and water can be represented as:
NH4+ + H2O ⇌ NH3 + H3O+
The equilibrium constant (Kb) for this reaction is given as 1.8 × 10^-5.
To find the pH of the NH4Cl solution, we can use the following steps:
Step 1: Write the expression for Kb:
Kb = [NH3][H3O+] / [NH4+]
Step 2: Since NH4Cl is a strong electrolyte, it fully ionizes into NH4+ and Cl- ions:
[NH4+] = 0.44 M
Step 3: Assume x as the concentration of NH3 and H3O+ ions formed. Since NH4Cl fully ionizes, the concentration of NH3 is equal to [NH4+] and the concentration of H3O+ is equal to [NH4+].
[NH3] = [H3O+] = x
Step 4: Substitute these values into the expression for Kb:
1.8 × 10^-5 = (x)(x) / (0.44 - x)
Step 5: Solve the quadratic equation for x.
1.8 × 10^-5 = x^2 / (0.44 - x)
x^2 = (1.8 × 10^-5) * (0.44 - x)
x^2 = (0.00792 - 1.8 × 10^-5x)
x^2 + 1.8 × 10^-5x - 0.00792 = 0
Using the quadratic formula, we find:
x = (-1.8 × 10^-5 ± √(1.8 × 10^-5)^2 - 4(1)(-0.00792)) / (2)
After calculating, we find x ≈ 0.00257 M.
Step 6: Calculate the pH using the concentration of H3O+:
pH = -log[H3O+]
pH = -log(0.00257)
pH ≈ 2.59
Therefore, the pH of the 0.44 M aqueous solution of NH4Cl is approximately 2.59.
To calculate the pH of a solution of NH4Cl, you need to consider the acidic nature of the ammonium chloride. The compound NH4Cl dissociates in water to form NH4+ (ammonium ion) and Cl- (chloride ion).
First, you need to identify the species that react with water. In this case, the NH4+ ion can act as an acid and donate a proton (H+) to water. The reaction is as follows:
NH4+ + H2O ⇌ NH3 + H3O+
The equilibrium constant for this reaction is called the acid dissociation constant, Ka. However, in this case, you are given the base dissociation constant, Kb, which is related to Ka through the autoionization constant of water (Kw).
The relationship between Ka, Kb, and Kw is as follows:
Kw = Ka x Kb
Given that Kb = 1.8 x 10^-5, you can calculate Ka as follows:
Ka = Kw / Kb = 1 x 10^-14 / (1.8 x 10^-5) = 5.56 x 10^-10
Now that you have Ka, you can proceed to calculate the concentration of H+ ions in the solution using the formula for acid dissociation:
[H+][NH3] / [NH4+] = Ka
Since the concentration of NH4Cl is 0.44 M, the concentrations of NH4+ and NH3 are both 0.44 M. Substituting the values into the equation:
[H+][0.44] / [0.44] = 5.56 x 10^-10
Simplifying the equation further:
[H+] = 5.56 x 10^-10
Now, to calculate the pH, which is the negative logarithm of the H+ concentration:
pH = -log[H+]
pH = -log(5.56 x 10^-10)
pH ≈ 9.3
Therefore, the pH of the 0.44 M aqueous solution of NH4Cl is approximately 9.3.