NH3 is a weak base (Kb = 1.8 × 10^–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.038 M in NH4Cl at 25 °C?
The Correct Answer and Explanation is:
To calculate the pH of a 0.038 M solution of NH4Cl, we first need to understand that NH4Cl dissociates in water to form NH4⁺ and Cl⁻ ions. The NH4⁺ ion is a weak acid, and it will hydrolyze in water to produce NH3 (a weak base) and H⁺ ions.
Step 1: Set up the equilibrium expression for NH4⁺ dissociation.
The dissociation of NH4⁺ in water can be represented by:
NH4++H2O⇌NH3+H3O+NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+NH4++H2O⇌NH3+H3O+
The equilibrium constant for this reaction is the Ka for NH4⁺. Since NH4Cl is the salt of NH3, we can find the Ka using the relationship between the Ka of NH4⁺ and the Kb of NH3:
Kw=Ka×KbK_w = K_a \times K_bKw=Ka×Kb
where K_w = 1×10−141 \times 10^{-14}1×10−14 at 25°C, and Kb=1.8×10−5K_b = 1.8 \times 10^{-5}Kb=1.8×10−5 for NH3. Rearranging, we can solve for Ka:
Ka=KwKb=1×10−141.8×10−5=5.56×10−10K_a = \frac{K_w}{K_b} = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}Ka=KbKw=1.8×10−51×10−14=5.56×10−10
Step 2: Set up an ICE table for the dissociation of NH4⁺.
The initial concentration of NH4Cl is 0.038 M, and since it dissociates 1:1, the initial concentration of NH4⁺ is also 0.038 M.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| NH4⁺ | 0.038 | -x | 0.038 – x |
| H⁺ (H3O⁺) | 0 | +x | x |
| NH3 | 0 | +x | x |
Using the Ka expression for the dissociation of NH4⁺:
Ka=[NH3][H3O+][NH4+]=x20.038−xK_a = \frac{[NH_3][H_3O^+]}{[NH_4^+]} = \frac{x^2}{0.038 – x}Ka=[NH4+][NH3][H3O+]=0.038−xx2
Since Ka is very small, we can assume that x will be much smaller than 0.038 M, so 0.038−x≈0.0380.038 – x \approx 0.0380.038−x≈0.038. This simplifies the equation to:
5.56×10−10=x20.0385.56 \times 10^{-10} = \frac{x^2}{0.038}5.56×10−10=0.038×2
Step 3: Solve for x.
Rearranging the equation to solve for x:
x2=(5.56×10−10)×0.038x^2 = (5.56 \times 10^{-10}) \times 0.038×2=(5.56×10−10)×0.038
x2=2.11×10−11x^2 = 2.11 \times 10^{-11}x2=2.11×10−11
Taking the square root of both sides:
x=2.11×10−11x = \sqrt{2.11 \times 10^{-11}}x=2.11×10−11
x=1.45×10−5x = 1.45 \times 10^{-5}x=1.45×10−5
This value of x represents the concentration of H⁺ ions in the solution.
Step 4: Calculate the pH.
The pH is given by:
pH=−log[H+]pH = -\log[H^+]pH=−log[H+]
Substituting the value of x:
pH=−log(1.45×10−5)pH = -\log(1.45 \times 10^{-5})pH=−log(1.45×10−5)
pH=4.84pH = 4.84pH=4.84
Therefore, the pH of the 0.038 M NH4Cl solution is 4.84.
Explanation:
The weak acid behavior of NH4⁺ in water leads to the formation of hydronium ions (H₃O⁺), lowering the pH below 7. The pH calculated here reflects the weak acidic nature of the NH4⁺ ion in solution.
