g Given that ammonia (NH3) acts as a weak base in water with a Kb of 1.8 x 10-5 at 298.0 K, calculate the pH at 298.0 K of a solution prepared by mixing 100.0 mL of a 0.050 M aqueous solution of ammonia with 20.0 mL of a 1.00 M aqueous solution of nitric acid. The final volume of the solution is 120.0 mL.

pH = 0.90

Explanation:

The NH3 reacts with nitric acid, HNO3 as follows:

NH3 + HNO3 → NH₄⁺ + NO₃⁻

Where 1 mole of NH3 reacts per mole of HNO3.

The moles added of each reactant are:

Moles NH3:

100mL = 0.100L * (0.050mol / L) = 0.0050 moles NH3

Moles HNO3:

20mL = 0.0200L * (1.00mol / L) = 0.0200 moles HNO3

That means HNO3 is the excess reactant and, after the reaction, its moles are:

0.0200mol - 0.0050mol = 0.0150 moles HNO3

In a final volume of 120mL = 0.120L, the molar concentration of HNO3 is:

0.0150 moles HNO3 / 0.120L = 0.125M HNO3

As nitric acid is a strong acid, [HNO3] = [H+]

[H+] = 0.125M

pH = -log[H+]

it would be d. henry moseley

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