Question

A buffer solution was prepared by adding 4.89 g of sodium acetate, $NaCH_3CO_2$, to $2.50 \times 10^2$ mL of 0.100 M acetic acid, $CH_3CO_2H$ ($K_a = 1.8 \times 10^{-5}$). a What is the pH of the buffer? $pH = 5.12$ Correct Find the pH using the Henderson-Hasselbalch equation: $[CH_3CO_2^-] = [NaCH_3CO_2] = \frac{4.89 \text{ g}}{0.250 \text{ L}} \times \frac{1 \text{ mol } NaCH_3CO_2}{82.03 \text{ g}} = 0.238 \text{ M}$ $pH = pK_a + \log\frac{[CH_3CO_2^-]}{[CH_3CO_2H]} = -\log(1.8 \times 10^{-5}) + \log\left(\frac{0.238}{0.100}\right) = 5.12$ b What is the pH of the system if you add 89 mg of NaOH to the $1.00 \times 10^2$ mL of the prepared buffer solution? $pH = $ Submit

          A buffer solution was prepared by adding 4.89 g of sodium acetate, $NaCH_3CO_2$, to $2.50 \times 10^2$ mL of 0.100 M acetic acid, $CH_3CO_2H$ ($K_a = 1.8 \times 10^{-5}$).
a What is the pH of the buffer?
$pH = 5.12$
Correct
Find the pH using the Henderson-Hasselbalch equation:
$[CH_3CO_2^-] = [NaCH_3CO_2] = \frac{4.89 \text{ g}}{0.250 \text{ L}} \times \frac{1 \text{ mol } NaCH_3CO_2}{82.03 \text{ g}} = 0.238 \text{ M}$
$pH = pK_a + \log\frac{[CH_3CO_2^-]}{[CH_3CO_2H]} = -\log(1.8 \times 10^{-5}) + \log\left(\frac{0.238}{0.100}\right) = 5.12$
b What is the pH of the system if you add 89 mg of NaOH to the $1.00 \times 10^2$ mL of the prepared buffer solution?
$pH = $ 
Submit

Show more…

A buffer solution was prepared by adding 4.89 g of sodium acetate, NaCH3CO2, to 2.50 × 10^2 mL of 0.100 M acetic acid, CH3CO2H (Ka = 1.8 × 10^-5).
a What is the pH of the buffer?
pH = 5.12
Correct
Find the pH using the Henderson-Hasselbalch equation:
[CH3CO2^-] = [NaCH3CO2] = (4.89 g)/(0.250 L)×(1 mol NaCH3CO2)/(82.03 g) = 0.238 M
pH = pKa + log([CH3CO2^-])/([CH3CO2H]) = -log(1.8 × 10^-5) + log((0.238)/(0.100)) = 5.12
b What is the pH of the system if you add 89 mg of NaOH to the 1.00 × 10^2 mL of the prepared buffer solution?
pH =
Submit

Added by Sarah H.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Supreeta N

04/19/2023

Step 1

[CH3COO-] = (mass of sodium acetate / molar mass of sodium acetate) / volume of solution [CH3COO-] = (4.89 g / 82.03 g/mol) / 0.250 L [CH3COO-] = 0.238 M Show more…

Show all steps

lock

Thanks for your feedback!

Part B A buffer solution was prepared by adding 4.89 g of sodium acetate, NaC2H3O2 to 2.50 x 10^2 mL of 0.100M acetic acid, CH3COOH. K=1.8 x 10^-5 a) What is the pH of the buffer? pH=5.12 Correct Find the pH using the Henderson-Hasselbalch equation: [CH3COO-]=[NaCH3COO-]= 0.250L 82.03 g [CH3COO-] pH=pKa+log (CH3COOH) 0.238 -log(1.8 x 10^-5)+log =5.12 0.100 pH= Submit

Play audio

Feedback

Powered by NumerAI

Supreeta N and 65 other subject Chemistry 101 educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 Hi, in this question we are given with buffer solution that has volume 100 ml and it consists of acetic acid and sodium acetate.

00:08 We are having the concentration of acetic acid and sodium acetate in this manner.

00:12 The pka value of acetic acid is 4 .76.

00:16 We are asked to calculate the ph of buffer solution.

00:19 Since it is an acetic buffer solution, therefore by using hederson hazelbrad's equation, we can say that the ph is equal to pka plus log concentration of salt that is sodium acetate over concentration of acid.

00:37 We can substitute the values here.

00:41 The pka value is 4 .76 plus log concentration of salt that is sodium acetate is 0 .110 molar divided by concentration of acid is 0 .100 molar.

00:54 Now on solving this we are getting the ph of buffer solution equals to 4 .80.

01:03 This is the answer of first part.

01:05 Now in second part of the question, we are given that 0 .150 molar of naoh is added to this buffer solution that has volume 10 ml which is equal to 0 .010 liter.

01:18 We are asked to calculate the ph of buffer solution on addition of this naoh.

01:23 First of all, we can write the reaction here.

01:26 Acetic acid reacts with naoh and it produces sodium acetate and water.

01:42 Due to addition of naoh, the moles of acetic acid decreases and moles of sodium acetate increases...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later