Question 2.
pH = pKa + Log{[CH3COONa - HCl]/[CH3COOH + HCl]}
Moles of CH3COONa = (100/1000) L * 1.5 mol/L = 0.15 mol
Moles of CH3COOH = (100/1000) L * 0.9 mol/L = 0.09 mol
Now, pH = 4.74 + Log{(0.15 - 0.03)/(0.09 + 0.03)}
Therefore, pH = 4.74
Question 3.
pH = pKa + Log{[CH3COONa + NaOH]/[CH3COOH - NaOH]}
Moles of CH3COONa = (200/1000) L * 1.5 mol/L = 0.3 mol
Moles of CH3COOH = (200/1000) L * 0.9 mol/L = 0.18 mol
Moles of NaOH = (150/1000) L * 0.02 mol/L = 0.003 mol
Now, pH = 4.74 + Log{(0.3 + 0.003)/(0.18 - 0.003)}
Therefore, pH = 4.97
Need help with number three please salt Find the pH of a buffer solution containing of...
Module 11 Homework 1. Find the pH of a buffer solution containing of 1.5M sodium acetate and 0.9M acetic acid. The K, of acetic acid is 1.8 x 105. 2. Find the pH of 100.0 mL of the buffer from exercise 1 after the addition of 0.030 mol of gaseous hydrochloric acid. Assume no volume change. 3. Find the pH of 200.0 mL buffer from exercise 1 after the addition of 150.0 mL of 0.020M NaOH. 4. What conjugate acid-base...
A 1.00 L buffer solution with pH = 4.74 is composed of 0.30 mol acetic acid and 0.30 mol sodium acetate. A) Determine the pKa of acetic acid B) If 0.030 mol of NaOH is added, determine the pH of the solution
=Assume you have prepared 100.0 mL of a buffer solution using 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate. The pH of this buffer solution is initially 4.74. After preparing this buffer solution, you added 55.0 mL of a 1.10 M NaOH solution to your buffer to see what would happen. What will the pH of this new solution be?
A 1.028 L buffer solution is 0.644 M in acetic acid and 0.644 M in sodium acetate. Acetic acid has a pKa of 4.74, making the pH of this solution 4.74. What is the pH after addition of 0.244 mol of HCl?
Create a buffer solution with a pH of 6.0. Based on the assigned pH, you will choose an appropriate buffer system. pH = pKa + log[A-}/[HA] A) 3.5M acetic acid 60.1MM, sodium acetate, anhydrous 82.0MM, pKa 4.74 OR B) sodium phosphate, monobasic, NaH2PO4*H2O 138.0MM, sodium phosphate, dibasic, Na2HPO4 142.0MM, pKa 6.86 Calculate the amount of weak acid and conjugate base you need to make 25mL of 0.25M solutions (one acid and one base solution).
Part A) you will be preparing a buffer solution. The solution is 50.0 mL of a 0.200 M acetate buffer, pH 5.00, starting from a 1.00 M solution of acetic acid and solid sodium acetate. pKa of acetic acid = 4.74 molar mass of sodium acetate = 82 g/mol Hint: Use the Henderson-Hasselbach equation (Eq. 21) Use the equation [salt] + [acid] = 0.200 M Please calculate the number of mL of 1.00 M acetic acid and the number of...
A buffer formed from 20.0 mL of 0.120 M Acetic acid (CH3COOH) (pKa = 4.74) and 12.0 mL of 0.200 M sodium acetate (CH3COONa). What is the pH of this buffer after addition of 0.0001 mol of HCl?
With this information how would I calculate the pH of
.01 M Ammonia solution and the pH of .01 M Ammonia buffer
solution
I need
Ph of acetic acid! sorry for the wrong wording!
A. Preparation of Acetic Acid-Acetate Buffer Solution An acetate buffer contains the acid-base pair, acetic acid and the acetate fon (typically added as sodium acetate). For acetic acid, pk, = -log (1.8 10 = 4.74. Consider the con ditions that will yield 100 ml of an...
(4 pts) What is the pH of the a buffer solution containing 0.10 moles of acetic acid and 0.12 moles of sodium acetate in 45 ml after adding 0.030 moles of KOH? 3. 0.049L Acenc acid + KOH CH3CO0K t H20 HOE
You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74.