NH4Cl is a weak base (Kb = 1.8 * 10^-5) in NH4Cl solution (0.035 M in NH4Cl) at 25°C, and so the salt NH4Cl acts as a weak acid. What is the pH of a solution with NH4Cl?
The Correct Answer and Explanation is:
To calculate the pH of a solution of NH4Cl, we need to consider how NH4Cl behaves in water. Since NH4Cl is the salt of a weak base (NH3) and a strong acid (HCl), it will dissociate into NH4+ and Cl-. The NH4+ ion will act as a weak acid and undergo hydrolysis in water, releasing H+ ions and lowering the pH.
Here’s the breakdown of how to find the pH:
- Hydrolysis Reaction of NH4+:
NH4+ + H2O ⇌ NH3 + H3O+ The NH4+ ion donates a proton to water, creating NH3 and hydronium ions (H3O+). The equilibrium constant for this reaction is the acid dissociation constant (Ka) for NH4+, which can be calculated from the given Kb for NH3. - Relationship between Ka and Kb:
Since NH4+ is the conjugate acid of NH3, we can use the relationship: Ka=KwKbK_a = \frac{K_w}{K_b}Ka=KbKw where K_w is the ionization constant of water (1.0 × 10^-14 at 25°C), and Kb is the given value for NH3, which is 1.8 × 10^-5. Therefore, Ka=1.0×10−141.8×10−5=5.56×10−10K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}Ka=1.8×10−51.0×10−14=5.56×10−10 - Set up the ICE Table:
For the hydrolysis reaction, let the initial concentration of NH4+ be 0.035 M. Let x be the concentration of H3O+ (which is also equal to the concentration of NH3 produced). The equilibrium concentrations will be:- Initial: [NH4+] = 0.035 M, [H3O+] = 0, [NH3] = 0
- Change: [NH4+] decreases by x, [H3O+] increases by x, [NH3] increases by x
- Equilibrium: [NH4+] = 0.035 – x, [H3O+] = x, [NH3] = x
- Expression for Ka:
The Ka expression is: Ka=[NH3][H3O+][NH4+]=x⋅x0.035−xK_a = \frac{[NH3][H3O+]}{[NH4+]} = \frac{x \cdot x}{0.035 – x}Ka=[NH4+][NH3][H3O+]=0.035−xx⋅x Since Ka is very small, we can approximate that 0.035 – x ≈ 0.035. So the equation simplifies to: 5.56×10−10=x20.0355.56 \times 10^{-10} = \frac{x^2}{0.035}5.56×10−10=0.035×2 Solving for x: x2=(5.56×10−10)⋅0.035=1.946×10−11x^2 = (5.56 \times 10^{-10}) \cdot 0.035 = 1.946 \times 10^{-11}x2=(5.56×10−10)⋅0.035=1.946×10−11 x=1.946×10−11=1.4×10−5 Mx = \sqrt{1.946 \times 10^{-11}} = 1.4 \times 10^{-5} \, \text{M}x=1.946×10−11=1.4×10−5M This is the concentration of H3O+ ions. - Calculate pH:
pH is calculated using the formula: pH=−log[H3O+]\text{pH} = -\log[H3O+]pH=−log[H3O+] pH=−log(1.4×10−5)=4.85\text{pH} = -\log(1.4 \times 10^{-5}) = 4.85pH=−log(1.4×10−5)=4.85
Thus, the pH of the 0.035 M NH4Cl solution is approximately 4.85.
This result indicates that the solution is slightly acidic, as expected from the hydrolysis of NH4+, the conjugate acid of a weak base.
