I have 50.00 mL of 0.100 M ethyl amine (C2H5NH2). I gradually add a solution of 0.025 M nitric acid (HNO3) to the ethyl amine solution. Required:
What is the pH after the addition of a total of 201 mL of the nitric acid?

4.00 is the pH of the mixture

Explanation:

The ethyl amine reacts with HNO3 as follows:

C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻

To solve this question we need to find the moles of ethyl amine and the moles of HNO3:

Moles C2H5NH2:

0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine

Moles HNO3:

0.201L * (0.025mol/L) = 0.005025 moles HNO3

That means HNO3 is in excess. The moles in excess are:

0.005025 moles HNO3 - 0.00500 moles ethyl amine =

2.5x10⁻⁵ moles HNO₃

In 50 + 201mL = 251mL = 0.251L:

2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]

As pH = -log [H+]

pH = -log 9.96x10⁻⁵M

pH = 4.00 is the pH of the mixture

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