What is the pH of a 0.20 M solution of NH4Cl? [Kb(NH3) = 1.8 × 10−5]
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To find the pH of a solution of NH4Cl, we need to consider the dissociation of NH4Cl into its ions. In this case, NH4Cl dissociates into NH4+ and Cl- ions.
The NH4+ ion is the conjugate acid of NH3 (ammonia) and can act as a weak acid in water. It will react with water to produce H3O+ ions and NH3:
NH4+ + H2O -> NH3 + H3O+
However, the Kb value of NH3 is given in the question, which indicates the base dissociation constant for NH3. To find the Ka value for NH4+, we can use the relationship:
Kw = Ka x Kb
Kw is the autoionization constant of water (1.0 x 10^-14), which is constant at a given temperature.
Therefore,
Ka = Kw / Kb
Ka = (1.0 x 10^-14) / (1.8 x 10^-5)
Now, we can calculate the concentration of OH- ions in the solution. Since NH4Cl is a strong electrolyte, it dissociates completely, and the concentration of NH4+ ions is equal to the initial concentration of the NH4Cl solution (0.20 M).
Next, we can use the Kb expression for NH4+ ions:
Kb = [NH3][OH-] / [NH4+]
Since [OH-] is small compared to the [NH4+] concentration, we can assume that the [OH-] generated by NH4+ is negligible compared to the OH- from water, so [OH-] ≈ 0. Therefore, we can rewrite the expression as:
Kb = [NH3] x 0 / [NH4+]
Simplifying further:
Kb = 0
It means that the Kb value is negligible compared to the NH4+ concentration. Thus, the [NH4+] concentration and hence the concentration of H3O+ ions are equal.
Therefore, pH = -log[H3O+]
pH = -log[0.20]
Using a calculator, we find:
pH ≈ 0.70
Thus, the pH of a 0.20 M solution of NH4Cl is approximately 0.70.
To determine the pH of a solution, we need to consider the dissociation of the compound in water. In this case, NH4Cl will dissociate into NH4+ (ammonium ion) and Cl- (chloride ion). The ammonium ion will react with water to produce NH3 (ammonia) and H3O+ (hydronium ion).
The reaction can be represented as follows:
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
Since NH4Cl is a strong electrolyte, it will readily dissociate completely in water. Therefore, we can assume that the initial concentration of NH4+ is 0.20 M, and the initial concentration of Cl- is also 0.20 M.
Now, we need to determine the concentration of NH3 (ammonia) and H3O+ (hydronium ion) in the solution. The concentration of NH3 can be calculated using the equilibrium constant Kb for the reaction between NH4+ and H2O. The Kb value given in the question is 1.8 × 10^-5, which represents the equilibrium constant expression:
Kb = [NH3][H3O+] / [NH4+]
Since we know the concentration of NH4+ (0.20 M), we can solve for the concentrations of NH3 and H3O+.
First, we assume that x is the concentration of NH3 and H3O+. This means that the concentration of NH4+ will decrease by x, while NH3 and H3O+ will increase by x.
Therefore, after the reaction reaches equilibrium:
[NH3] = [H3O+] = x
[NH4+] = 0.20 - x (since the initial concentration of NH4+ is 0.20 M and it decreases by x)
Now, we can substitute these values into the Kb expression:
1.8 × 10^-5 = (x)(x) / (0.20 - x)
Since the value of x is small compared to 0.20, we can assume that 0.20 - x is approximately 0.20.
1.8 × 10^-5 = (x)(x) / 0.20
Simplifying, we get:
x^2 = 1.8 × 10^-5 * 0.20
x^2 = 3.6 × 10^-6
x = √(3.6 × 10^-6)
x ≈ 1.897 × 10^-3
So, the concentration of NH3 and H3O+ is approximately 1.897 × 10^-3 M.
Next, we need to calculate the pH. The pH is a measure of the concentration of H3O+ in a solution. It is defined as the negative logarithm (base 10) of the H3O+ concentration.
To find the pH, we can use the formula:
pH = -log[H3O+]
Substituting the concentration of H3O+ (1.897 × 10^-3 M), we get:
pH = -log(1.897 × 10^-3)
Calculating this on a calculator, we find that:
pH ≈ 2.72
So, the pH of the 0.20 M solution of NH4Cl is approximately 2.72.
NH4^+ + H2O ==> H3O^+ + NH3
Set up an ICE chart, then substitute into the following:
Ka = (Kw/Kb) = (H3O^+)(NH3)/(NH4^+)
(H3O^+) = x = (NH3)
(NH4^+) = (0.2-x)