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A 100.0 mL sample of 0.10 M NH3 (weak base) is titrated with 0.10 M HNO3 (strong acid). Determine the pH of the solution after the addition of 50.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.

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To solve this problem, we need to use the concept of weak acid-base equilibrium and the Henderson-Hasselbalch equation. The reaction between NH3 and HNO3 can be written as:

NH3 + HNO3 → NH4+ + NO3-

Before any HNO3 is added, the NH3 solution is a weak base with the following equilibrium equation:

NH3 + H2O ⇌ NH4+ + OH-

where the Kb of NH3 is 1.8 × 10-5. At the start, we have 0.10 M NH3, and the concentration of OH- can be calculated using the Kb expression:

Kb = [NH4+][OH-]/[NH3]

[OH-] = Kb[NH3]/[NH4+] = 1.8 × 10^-5 × 0.10 / x, where x is the concentration of NH4+.

At the equivalence point, the moles of HNO3 added equals the moles of NH3 initially present, and the solution contains only NH4+ and NO3-. Therefore, the concentration of NH4+ is:

x = 0.10 - 0.05 = 0.05 M

At this point, all the OH- ions have been consumed by the HNO3, so the pH of the solution depends on the concentration of NH4+. The Henderson-Hasselbalch equation for a weak acid-base system is:

pH = pKa + log([base]/[acid])

where pKa is the negative logarithm of the acid dissociation constant (Ka) and [base]/[acid] is the ratio of the concentrations of the weak base and its conjugate acid.

For NH3, the conjugate acid is NH4+ and the pKa can be calculated as:

pKa = -log(Ka) = -log(1.8 × 10^-5) = 4.74

Plugging in the values, we get:

pH = 4.74 + log(0.05/0.05) = 4.74

Therefore, the pH of the solution after the addition of 50.0 mL of HNO3 is 4.74.

User Pustovalov Dmitry
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