In a solution of 0.1 M NH4Cl, what is the pH if the Kb of NH3 is 1.8 x 10^-5?

In a solution of 0.1 M NH4Cl, what is the pH if the Kb of NH3 is 1.8 x 10^-5?

Step 1: Identify the concentration of the NH4Cl solution, which is 0.1 M.
Step 2: Recognize that NH4Cl is a salt that comes from the weak base NH3 and strong acid HCl.
Step 3: Understand that NH4+ (from NH4Cl) will act as a weak acid in solution.
Step 4: Use the relationship between Kb (for NH3) and Ka (for NH4+) using the formula: Ka = Kw / Kb, where Kw is 1.0 x 10^-14.
Step 5: Calculate Ka: Ka = (1.0 x 10^-14) / (1.8 x 10^-5) = 5.56 x 10^-10.
Step 6: Set up the expression for the dissociation of NH4+: NH4+ ⇌ H+ + NH3.
Step 7: Write the expression for Ka: Ka = [H+][NH3] / [NH4+].
Step 8: Assume x is the concentration of H+ produced, so [H+] = x, [NH3] = x, and [NH4+] = 0.1 - x (approximately 0.1 since x is small).
Step 9: Substitute into the Ka expression: 5.56 x 10^-10 = (x)(x) / (0.1).
Step 10: Rearrange to find x: x^2 = (5.56 x 10^-10)(0.1) = 5.56 x 10^-11.
Step 11: Solve for x: x = sqrt(5.56 x 10^-11) = 7.45 x 10^-6.
Step 12: This x value represents [H+], so calculate pH: pH = -log(7.45 x 10^-6).
Step 13: Calculate pH: pH ≈ 5.13.
Step 14: Alternatively, use the shortcut: pH = 14 - pKb, where pKb = -log(1.8 x 10^-5) ≈ 4.75, so pH = 14 - 4.75 = 9.25.

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