Hi There! I would like some help with the following question : A student is aske

Question

Hi There! I would like some help with the following question :
A student is asked to prepare 100 ml of 0.02M CH3COONa/ 0.02M CH3COOH buffer.

(a) What is the mass of CH3COONa and CH3COOH he must use?
My answer : 0.164g and 0.12g respectively.

(b) The ionisation of CH3COOH in water yields hydronium (H3O) and CH3COO. Write the reversible equation for the dissociation and identify Bronsted acid and base as well as conjugate acid and base.

My answer : CH3COOH (Bronsted acid) + H2O (base) <-> CH3COO (Bronsted Base) + H3O (conjugate acid)

(c) Calculate the pH of the buffer solution. Assume dissociation to be very small. Ka of Ch3COOH is 1.8 x 10^-5

My answer : ph = -log 0.0006 = 3.22

(d) What is the volume of 0.1M HCl he must add to reduce the pH to 4.70?

I do not know how to do this.

(e) What can be done to increase buffering capacity?

I do not know how to do this.

Much appreciated, thank you!

A student is asked to prepare 100 ml buffer to prepare 100 ml of 0.02M CH,COONa /0.02M CH,COOH (#y Calculate the mass of CH,COONa and CH3COOH he must use. (4 marks) (o) The ionisation of CHicooH in water yields hydronium (H,0) and CH3COO. Write the reversible equation for the dissociation and identify (3 marks) ) Calculate the pH of the buffer solution. You may assume dissociation to (6 marks d) What is the volume of 0.1M HCI he must add if he wants to reduce th (6 mark the Bronsted acid and base, as well as the conjugate acid and base. be very small. The Ka of CH,COOH is 1.8 x 10. pH to 4.702 ) What can he do to increase the buffering capacity? (2 mark iven that E-0.52 V for the reduction Cu+(aq) + e Cu(s), calculate E following reaction at 25 °C: 2Cu+(aq) Cu2+(aq) + Cu(s) (5 mar galvanic cell consists of a silver electrode in contact with 346 mL of 0.11 NO3 aqueous solution and a magnesium electrode in contact with 288 ml 0 M Mg(NO3)2 aqueous solution. Calculate E for the cell at 25 °C. (5 ma A current is drawn from the cell until 1.20 g of silver has been depo at the silver electrode. Calculate E for the cell at this stage of operat (5 m

Explanation / Answer

CH3COOH is a weak acid - calculate [H+] from the Ka equation for question a)
When you add NaOH to the acid you produce a buffer solution Use the Henderson Hasselbalch equation
At the half equivalence point , pH = pKa . pKa for the acid is 4.77 . The half equivalence point is c) So the pH here will be 4.77

I will leave you to rework the problem . If you have not made any progress , or someone else has not done the work for you , I will come back to this problem tomorrow.
Edit next day:
No progress from you and no other answer: So we will do this in full
Question a) You have a weak acid. In order to calculate pH you first have to determine the [H+] in the solution. You do this using the Ka equation:

Ka = [H+] [CH3COO-] / [CH3COOH]
You know that [H+] = [CH3COO-] so for product we write [H+]²
Because the acid is weak , we say [CH3COOH] = 0.10
Substitute:
1.7*10^-5 = [H+]² / 0.1
[H+]² = (1.7*10^-5) *0.1
[H+]² = 1.7*10^-6
[H+] = (1.7*10^-6)
[H+] = 1.30*10^-3
pH = -log [H+]
pH = -log (1.3*10^-3)
pH = 2.88

Now we go to b) but first some theory:
When you add NaOH to the CH3COOH solution you form CH3COONa. You then have a solution of a weak acid and a salt of this acid. This is a buffer solution . The pH of the buffer solution is calculated using the Henderson - Hasselbalch equation.
pH = pKa + log ([salt] /[acid])
Some preliminary calculations :
pKa = -log Ka = - log (1.7*10^-5) = 4.77

Mol CH3COOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol
Mol NaOH in 25mL of 0.1M solution = 25/1000*0.1 = 0.0025 mol
The NaOH reacts with the CH3COOH in 1:1 ratio
Therefore you produce 0.0025 mol CH3COONa
and remaining unreacted is 0.0075 mol CH3COOH
In total volume 125mL = 0.125 L
Molarity of CH3COONa = 0.0025/0.125 = 0.02M
Molarity of CH3COOH = 0.0075 / 0.125 = 0.06M
Now use the H-H equation:
pH = pKa + log ([salt] /[acid])
pH = 4.77 + log ( (0.02/0.06)
pH = 4.77 + log 0.333
pH = 4.77 + (-0.48)
pH = 4.29

c) This is the really interesting situation : remember I said that here pH = pKa. Here's how:
Mol CH3COOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol
Mol NaOH in 50mL of 0.1M solution = 25/1000*0.1 = 0.005 mol
The NaOH reacts with the CH3COOH in 1:1 ratio
Therefore you produce 0.005 mol CH3COONa
and remaining unreacted is 0.005 mol CH3COOH
In total volume 150mL = 0.150 L
Molarity of CH3COONa = 0.005/0.150 = 0.0333M
Molarity of CH3COOH = 0.005 / 0.150 = 0.0333M
Substitute into H-H equation:
pH = 4.77 + log (0.0333*0.0333)
pH = 4.77 + log 1
pH = 4.77 + 0
pH = 4.77 which is the same as pKa

d) I am sure that you can do this by yourself: Follow the above system

e) When you have added 100mL of the NaOH solution you have neutralised all the acid. There is then only a solution of salt CH3COONa in solution . This is called the equivalence point - and you must calculate the pH at the equivalence point.

You have reacted 0.01mol CH3COOH with 0.01mol NaOH to produce 0.01 mol CH3COONa dissolved in 200mL = 0.20L solution:
Molarity of CH3COONa solution = 0.01/0.2 = 0.05M solution.
Calculate pH as follows:
The aim is to find the [OH-] of the solution You will know that a salt of a strong base and weak acid is basic.
The CH3COONa dissociates in water :
CH3COONa CH3COO- + Na+ The CH3COO- reacts with water:
CH3COO- + H2O CH3COOH + OH-

Ka for CH3COOH = 1.7*10^-5
Kw = Ka * Kb - we want Kb
Kb = Kw* Ka
Kb = 10^-14 / (1.7*10^-5)
Kb = 5.88*10^-10

Kb = [CH3COO-] *[[OH] / [CH3COONa]
We know that [CH3COO-] = [OH-] so product is [OH-]²
[CH3COONa] = 0.05M

Kb = [OH]²/0.05
(5.88*10^-10 ) * 0.05 = [OH[²
[OH]² = 2.94*10^-11
[OH-] = 5.42*10^-6
To calculate pH you require [H+]
[OH-] *[H+] = 10^-14
[H+] = 10^-14/ [OH-]
[H+] = 10^-14 / (5.42*10^-6)
[H+] = 1.84*10^-9

pH = -log ( 1.84*10^-9)
pH = 8.73

And finally to f) You have added a very large excess of NaOH . All the acid has been neutralised. With this large excess of a strong base , the salt does not play any role in the pH of the final solution.
You have added 300mL of NaOH = 0.03 mol NaOH
This has reacted with 100mL = 0.01mol acid.
You have 0.02 mol NaOH remaining unreacted in a final volume of 400mL
Molarity of NaOH solution: 0.02/0.4 = 0.05 M NaOH

In 0.05M NaOH solution , [OH-] = 0.05M
[H+] = 10^-14 / 0.05
[H+] = 2.0*10^-13

pH = -log 2.0*10^-13
pH = 12.7

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