Multiple Choice
Calculate the pH of a solution made by adding 160 mL of 1.00 M HCl to 300 mL of 1.70 M NH3 solution.
17. Acid and Base Equilibrium
pH of Weak Bases
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What is the pH of a 0.208 M NH4Cl solution at 25°C, given that the Kb of NH3 is 1.8x10^-5?
A
4.98
B
7.00
C
9.02
D
11.22
Verified step by step guidance1
Identify that NH4Cl is a salt formed from the weak base NH3 and the strong acid HCl. In solution, NH4+ will act as a weak acid.
Write the hydrolysis reaction for NH4+: NH4+ + H2O ⇌ NH3 + H3O+. This shows NH4+ donating a proton to water, forming NH3 and H3O+.
Use the relationship between Ka and Kb for conjugate acid-base pairs: Ka * Kb = Kw, where Kw is the ion-product constant of water (1.0 x 10^-14 at 25°C). Calculate Ka for NH4+ using Ka = Kw / Kb.
Set up the expression for the acid dissociation constant (Ka) of NH4+: Ka = [NH3][H3O+] / [NH4+]. Assume [NH3] = [H3O+] = x and [NH4+] = 0.208 - x, but since x is small, approximate [NH4+] ≈ 0.208.
Solve for x, which represents [H3O+], using the equation Ka = x^2 / 0.208. Once x is found, calculate the pH using pH = -log[H3O+].
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