Chapter 9: Unit 10. Buffers

Buffers

Buffers are solutions that resist a change of the pH and maintaining the pH of such solutions unchanged by neutralizing small amounts of added acids or bases.

Examples of such buffers solutions are:

• a. Human blood
• b. NH3 / NH4 +
• c. HCO3 – / CO32–
• d. CH3COOH / CH3COO–
• e. H3PO4 / H2PO4–

From the above examples, one can deduce that the buffer solutions are made of:

• 1. Weak Acid Mixed with Its Conjugate Base
• 2. Weak Base Mixed with its Conjugate Acid

Note: Strong Acids and Strong Bases can NOT form buffer solutionsbecause they cannotbe hydrolyzed in aqueous solutions.

A video of you tube illustrates the concept of the buffer solutions:

Let us consider the buffer solution system of NH 3 / NH 4 Cl

NH 3 (aq) + H 2 O(l) ⇔NH 4 + (aq) + OH–(aq)

The Buffer System is made of NH3/ NH4 +

Now let us examine the buffer function:

If an external acid [H 3 O + ] is added to this buffer solution, then the external acid [H 3 O + ] will react with [OH–] of the buffer solution to produce water:

H3O + (aq) + OH(aq)– ⇔H2O(aq)

When OH– is removed by H3O + , the equilibrium will shift to the right to the products’ side of the equilibrium. This means the amount of NH 4 + will slightly be increased on the expense of NH 3 which will be slightly decreased.

On the other hand, when an external base [OH–] is added to the buffer solution, then the external base [OH–] will cause the equilibrium to shift to the side of the reactants.This means that the amount of NH 3 will slightly be increased on the expense of NH 4 + which will be slightly decreased.

The increase and the decrease of the concentrations of NH3 and NH4 + is very small, thus the pH of the buffer solution is maintained.

pH Calculations of the Buffer Solutions

pH and Buffers.mp4

https://youtu.be/7Us44X98r-E

Example ofK aof Formic Acid

Let us consider the buffer solution: HCOOH / HCOONa [Formic Acid / Sodium Formate]:

HCOOH(aq) + H 2 O(l)‌⇔ HCOO–(aq) + H3O + (aq) [with K a = 1.80 x 10–4 ]

Setting up the Ka expression:

K a = { [HCOO–(aq)] x [H3 O+ (aq)] } / [HCOOH(aq)]

Isolating [H3 O + (aq)]:

[H3O + (aq)] = { K a x [HCOOH(aq)] } / [HCOO–(aq)]

[H3O + (aq)] = K a x { [HCOOH(aq)] } / [HCOO–(aq)] }

Since the pH is defined as – Log [ H3O + ], then taking the Log of both sides of the equations, one obtains the final equation of the pH of the buffer solution:

–Log [H3O + (aq)] = –Log [ K a x { [HCOOH(aq)] } / [HCOO–(aq)] } ]

–Log [H3O + (aq)] = –Log K a – Log { [HCOOH(aq)] } / [HCOO–(aq)] } ]

pH = pK a+ Log [HCOO–(aq)] / [HCOOH(aq)]

The equationin red aboveis known as Henderson –Hasselbalch Equation and it is used to calculate the pH of the buffer solutions.

Let us look at simple calculations of the pH of some buffer solutions:

Example:

What is the pH of a buffer solution that is 0.25 M in HF and 0.10 M in NaF? (K a for HF is 6. 8 x10–4 )

Solution:

pH = pK a + Log [F– (aq)] / [HF(aq)]

pH = –Log(6. 8 x10–4 ) + Log (0.10 M / 0.25)

pH = 3.16749 – 0.39794 = 2.76955 = 2.77[ 2 digits after decimal]

Example:

Determine the pH of a buffer that is 0.250 M in NH 3 and 0.500 M in NH4CI.

(Kb for NH 3 is 1.79 x 10–5 )

NH 3 (aq) + H 2 O(l) ß à NH 4 + (aq) + OH–(aq) [ Kb = 1.79 x 10–5 ]

K b = 1.79 x 10–5 = {[NH 4 + (aq)] x [OH–(aq)]} / [NH 3 (aq)]

Isolating [OH–(aq)]:

[OH–(aq)] = { (1.79 x 10–5 ) x [[NH 3 (aq)] } / [NH 4 + (aq)]

[OH–(aq)] = { (1.79 x 10–5 ) x (0.250 M)} / (0.500 M) = 8.95 x 10–6

K w = 1.00 x 10–14 = [OH–] x [H 3 O + ] = [8.95 x 10–6 ] x [H 3 O + ]

[H 3 O + ] = (1.00 x 10–14 ) / (8.95 x 10–6 ) = 1.12 x 10–9

pH = –Log (1.12 x 10–9 ) = 8.950 ( 3 digits after decimal)

Example:

Abuffer solution of pH 8.50 is desired.

Starting with 0.0100 mol of KCN and the usual inorganic reagents of the laboratory. How would you prepare 1.00 L of the buffer solution? ( Ka = 4.93 x 10–10)

The buffer solution will be made of the weak acid HCN and its conjugate base KCN (CN–):

HCN(aq) + H2O(l) ß à H3O+(aq) + CN–(aq)

pH = pKa + Log {[ CN–(aq)] / [HCN(aq)]

8.50 = (–Log(4.93 x 10–10) + Log [(0.0100 mol/1.00 L) / [HCN]]

8.50 = 9.307 + Log (0.0100 M) / [HCN]

8.50 –9.307 = – 0.807 = Log (0.0100 M) / [HCN]

– 0.807 = – Log { [HCN] / (0.0100 M) }

0.807 = Log { [HCN] / (0.0100 M) }

Taking the anti Log of both sides:

10.807 = { [HCN] / (0.0100 M) }

6.412 = { [HCN] / (0.0100 M) }

[HCN] = 6.412 x 0.0100 = 0.0641M

Both concentrations of the buffer solutions are known:

[HCN] = 0.0641M and [ CN–] = 0.0100 M

Now we have to convert the molarities into grams which can be weighed experimentally:

[HCN] = 0.0641mol/L

The molar mass of HCN = 1H + 1C + 1N = (1×1.00g/mol) + (1×12.0 g/mol) + (1×14.0 g/mol) = 27.0 g/mol

The amount of the buffer solution to be prepared is 1.00 L

Amount of grams of HCN:

[0.0641mol HCN / L HCN] x [27.0 g HCN / mol HCN] x [1.00 L HCN] = 1.73 g HCN

Amount of grams of KCN:

[ CN–] = [KCN] = 0.0100 mol/L

Molar mass of KCN = 1K + 1C + 1N = (1×39.0g/mol) + (1×12.0g/mol) + (1×14.0g/mol) = 65.0 g/mol

[0.0100 mol KCN / L KCN] x [65.0 g KCN / mol KCN] x [1.00 L KCN] = 0.650 g KCN

The buffer solution is made by weighing 1.73 grams of HCN and 0.650 grams KCN in 1.00 Liter H2O

Worksheet based on Simulation:

What is a buffer?

A buffer is a solution mixture made of:

1. A weak acid with its salt (conjugated base)

Example:

CH₃COOH (aq)  +  H₂O (l)   ßà    H₃O ⁺ (aq)  +  CH₃COO ^– (aq)

Acid Dissociation Constant = Ka = 1.76 × 10⁻⁵, the pKa = 4.76

Acetic acid    +     Water       ßà    Hydronium ion     + acetate ion

A mixture of acetic acid and sodium acetate will make a buffer

Or

1. A weak base with its salt (conjugated acid)

Example:

NH₄OH (aq)   ßà    NH₄ ⁺ (aq)     +    OH ^– (aq)

Base Dissociation Constant = Kb = 1.76 × 10⁻⁵, the pKb = 4.76

Ammonium hydroxide ßà Ammonium ion   + hydroxide ion

A mixture of ammonium hydroxide and ammonium chloride will make a buffer

The buffer is a solution that resists the change in the pH of the solution. It keeps the pH of the solution not changed (or very minor change).

If a small amount of an acid or a base is added to the buffer solution, the buffer will resist any change of the pH of the mixture solution.

Examples of Buffer solutions:

Acidic Buffers:

 CH₃COOH / CH₃COONa

 H₂CO₃ / NaHCO₃

 H₃PO₄ / NaH₂PO₄

 HCOOH / HCOONa

Basic Buffers:

NH₄OH / NH₄Cl

NH₃ / NH₄Cl

NH₃ / (NH₄)₂CO₃

The Henderson – Hasselbalch equation is used to calculate the pH of the solution

Henderson – Hasselbalch equation:

pH =  pKa    +    log     { [ A− ]   / [ HA ] }

Where pH = – log [ H ⁺ ] = – log [ H₃O ⁺ ]    à measured the acidity and the basicity of an aqueous solution

In our experiment:

[A−] = conjugated base = sodium acetate = NaCH₃COO, actually CH₃COO ^– (acetate) is the conjugated base. Na ⁺  is spectator ion and has no effect.

[ HA ] = weak acid = acetic acid

We will make a buffer of a solution by mixing acetic acid with sodium acetate.

Then we will test the buffer solution by adding externally HCl hydrochloric acid.

If the buffer is made correctly, addition small amount of HCl should not change the pH that much.

Before logging into the simulation, please watch this video about the simulation itself at the link below:

http://www.chemcollective.org/chem/common/vlab_walkthrouh_html5.php

After watching the video, you are ready, you are ready to start your lab.

Please follow all steps. Please do not skip any step.

1. Please log into the Buffer simulation by clicking on the link below:

http://chemcollective.org/vlab/104

2. Click on File to select the experiment:

Select: Load an Assignment and select: Acids and Bases:

4. Select the assignment: Creation of Buffer Problem by the clicking on the small arrow at right:

5. Click on the Buffer Creation Problem at the top right of the simulation:

6. A window will pop up with the information:

7. This is the lab assignment for the lab which means that you will have to create a buffer made of weak acid (acetic acid) with known amount and concentration mixed with the conjugated base (sodium acetate) with known amount and concentration. The pH of this created buffer has to be calculated using the Henderson – Hasselbalch equation.

8. Then you will have to add a small amount of a strong acid HCl to the buffer mixture and watch how the pH is changing. You will record you’re the amount of added HCl and calculate the pH

9. You keep adding acid to break the buffer capacity and ability of resisting the change of the pH and you record your data in a given table.

10. Make sure now that the solutions icon is selected:

11. Click on NaCH₃COO sodium acetate with the molarity of 1.0 M and volume of 100.0 mL at 25 ^o You will have to scroll all the way to find sodium acetate. Then click on stockroom to go back the solutions icon and scroll up and click on CH₃COOH acetic acid with the molarity of 1.0 M and volume of 100.0 mL at 25 ^oC. Record their original pH values. See the figures below:

12. Now put the sodium acetate Erlenmeyer flask exactly on the top of acetic acid Erlenmeyer flask. As soon you put both Erlenmeyer flasks on each other, then you will be ready to add the sodium acetate to the acetic acid solution.

Now, you can start your addition of sodium acetate 1 mL stepwise and record the pH.

Use the button: Pour

Stop the addition when the pH has reached the value of 4.76. Then record the amount of the sodium acetate inserted into the acetic acid. Also record the amount of the acetic acid and sodium acetate mixture.

13. In case you want to correct and repeat your addition of sodium acetate to acetic acid, click the X to reset the addition process and you start over.
14. You will need to put the 10 M HCl Erlenmeyer flask on the top of buffer solution mixture of acetic acid and sodium acetate to start the addition.
15. Now start adding 1 mL of 10 M HCl. Does the pH change dramatically? Did you create a buffer solution? Yes or No.

16. Keep adding HCl at rate of 1 ml. Record the amount of mL’s is needed to break the capacity and the ability of the buffer to resist any change of the pH. How many mL’s of 10 M HCl needed to reach the pH of 3.76.

[to read more about breaking the buffer ability and capacity to resist the change of the pH, please read the article in the same website with link:

http://collective.chem.cmu.edu/buffers/buffers5.php

17. Fill your data and findings so far in the given tables below:

Sodium acetate and acetic acid data:

Solution	Molarity	Volume	Temperature	pH
Sodium acetate
Acetic acid

Adding sodium acetate to acetic acid data: stepwise addition of 1 mL of sodium acetate to 100 mL of acetic acid till the pH 4.76

Solution	Total Volume added to acetic acid	Desired pH = 4.76
Sodium acetate

Solution	Total Volume of both acetic acid and sodium acetate	Desired pH = 4.76
Acetic Acid+ Sodium Acetate

18. Now let us see how to calculate the pH of the buffer using Henderson – Hasselbalch equation.

It is simple calculation.

pH=  pKa    +    log     { [ A− ]   / [ HA ] }

HA    ßà    A ^–    +    H ⁺       

In our example:

CH₃COOH (aq)  +  H₂O (l)   ßà    H₃O ⁺ (aq)  +  CH₃COO ^– (aq)

[HA] = [CH₃COOH]

[H ⁺]  =   [H₃O ⁺]

[A ^–] =   [CH₃COO ^–]

Ka  = 1.76 × 10⁻⁵,

pKa =  – log [1.76 × 10⁻⁵]   =  4.76

p   = – log

Now calculate the molarity of each of sodium acetate

Example:

           [HA] = [CH₃COOH] :  5M and 50 mL used for the buffer mixture

          [H ⁺]  =   [H₃O ⁺]

          [A ^–] =   [CH₃COO ^–] :  5M and 50 mL used for the buffer mixture

          Ka  = 1.76 × 10⁻⁵,         

Concentration of the acetic acid after the mixing = (5 mol / L) x 0.05 L = 0.25 mol

         [Note that we converted the 50 mL of the acid into Liters and then multiplied it by the    

        molarity of the acid]

        Concentration of the sodium acetate after the mixing = (5 mol / L) x 0.05 L = 0.25 mol

         [Note that we converted the 50 mL of the acetate into Liters and then multiplied it by the   

        molarity of the sodium acetate]

Total volume of both solutions: acetic acid and sodium acetate is = 50 mL + 50 mL = 100 

        mL = 0.1 Liter

Concentration of acetic acid after mixing both solutions: 0.25 mol / 0.1 L = 2.5 mol/L= 2.5 M

Concentration of sodium acetate after mixing both solutions: 0.25 mol / 0.1 L = 2.5 mol/L= 2.5 M

Now calculate the actual acidic acid and sodium acetate solution mixture used in this actual simulation:

Solution	MolarityBefore the Mixing	Volume Before the Mixing	Total Volume after the Mixing	MolarityAfter the Mixing
Sodium acetate
Acetic acid

Now let examine the addition of HCl to the buffer solution.

We will use the same example above:

CH₃COOH (aq)  +  H₂O (l)   ßà    H₃O ⁺ (aq)  +  CH₃COO ^– (aq)

After the mixing acetic acid with sodium acetate:

[HA] = [CH₃COOH] = 2.5 M

[H ⁺]  =   [H₃O ⁺]

[A ^–] =   [CH₃COO ^–] = 2.5 M

Ka  = 1.76 × 10⁻⁵,

pKa =  – log [1.76 × 10⁻⁵]   =  4.7

After adding 1 mL, 5 M HCl

Then the total the volume of the mixture (HCl+ CH₃COOH+NaCH₃COO) will be = 50 + 50 +1 = 101 mL = 0.101 L

[HA] = [CH₃COOH] = (2.5 mol /L) x (0.101 L) = 0.2525 mol

[A ^–] =   [CH₃COO ^–] = (2.5 mol /L) x (0.101 L) = 0.2525 mol

[HCl] = (5 mol /L) x (0.101 L) = 0.505 mol

Now calculate the actual acidic acid and sodium acetate solution mixture used in this actual simulation with adding 1 mL HCl

Solution	MolarityBefore the Mixing	Volume Before the Mixing	Total Volume after the Mixing	MolarityAfter the Mixing
Sodium acetate
Acetic acid
HCl

This section is NOT required. Just look at it and see how to calculate the pH using the

Henderson – Hasselbalch equation:

Without adding HCl

pH =  pKa    +    log     { [ A− ]   / [ HA ] }

pH = (-log 1.76 x 10^-5)  + log { 0.2525 / [ 0.2525] }

pH  = 4.76  +  0  = 4.76

After adding HCl:

HCl concentration will be added to the acetic acid concentration

HCl concentration will be subtracted from the sodium acetate concentration

pH =  pKa    +    log     { [ A− – HCl ]   / [ HA + HCl ] }

pH = (-log 1.76 x 10^-5)  + log { (0.505 – 0.2525) /  (0.505 + 0.2525) }

pH  = 4.76  +  log 0.333  = 4.76 – 0.477 = 4.283