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Multiple Choice
What is the pH of a 0.16 M NH4Cl solution, given that the Kb for NH3 is 1.76 × 10⁻⁵?
A
7.00
B
11.25
C
9.02
D
4.98
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Verified step by step guidance
1
Identify that NH4Cl is a salt formed from the weak base NH3 and the strong acid HCl. In solution, NH4Cl dissociates into NH4+ and Cl-. The NH4+ ion can donate a proton to water, acting as a weak acid.
Write the equilibrium expression for the hydrolysis of NH4+: NH4+ + H2O ⇌ NH3 + H3O+. The equilibrium constant for this reaction is the acid dissociation constant, Ka, which can be found using the relation Ka = Kw / Kb, where Kw is the ion-product constant of water (1.0 × 10⁻¹⁴ at 25°C).
Calculate Ka using the given Kb for NH3: Ka = (1.0 × 10⁻¹⁴) / (1.76 × 10⁻⁵).
Set up the expression for the equilibrium concentrations using the initial concentration of NH4Cl (0.16 M) and the change in concentration (x) due to the hydrolysis: [NH4+] = 0.16 - x, [NH3] = x, [H3O+] = x. Substitute these into the expression for Ka: Ka = [NH3][H3O+] / [NH4+].
Assume x is small compared to 0.16 M, simplifying the expression to Ka ≈ x² / 0.16. Solve for x, which represents the [H3O+], and then calculate the pH using the formula pH = -log[H3O+].