Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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?

Sagot :

jufsanabriasa

Answer:

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

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.