Question

Lab 5: Buffers Data analysis 1. Write reaction equations to explain how your acetic acid- acetate buffer reacts with an acid and with a base, respectively. 2. Buffer capacity has a rather loose definition, yet it is an important property of buffers. A commonly seen definition of buffer capacity is: "The amount of Hor OH that can be neutralized before the pH changes by one unit." Use your data to determine the buffer capacity of Buffer A and Buffer B. 3. Say, for example, that you had prepared a Buffer C, in which you mixed 8.203 g of sodium acetate, CH,COONa, with 100.0 mL of 1.0 M acetic acid. What would be the initial pH of Buffer C? 4. If you add 5.0 mL of 0.5 M NaOH solution to 20.0 mL each of Buffer B and Buffer C, which buffer's pH would change less? Explain.

Answer 1

Answer 1

Acetic acid –acetate buffer is made by mixing acetic acid (CH₃COOH) and sodium acetate (CH₃COONa). in aqueous medium these two ionize as given below:

CH₃COOH    H⁺ + CH₃COO^–

CH₃COONa Na⁺ + CH₃COO^–

Acetic acid is a weak acid and weakly ionizable. Here you can see a common ion effect which further suppress the ionization of acetic acid. When an acid is added to this buffer it provides H+ ion which react with acetate ion (CH₃COO^–) and form acetic acid. As the ionization of acetic acid has already suppressed due to common ion effect, the H+ ion concentration remain same and there is no change in pH.

H⁺ + CH₃COO^–           CH₃COOH      

If a base is added it provide hydroxyl ion (OH^-) which reacts with (H⁺) to give water molecule and hence there is no change in pH.

H⁺ + OH^–           H₂O      

Answer 3

Given

Mass of CH₃COONa = 8.203 gm

Molarity of CH₃COOH = 1.0 M

Volume of buffer =100 ml

First we will calculate molarity of CH₃COONa

molarity of CH₃COONa=mass of CH₃COONa molar mass of CH₃COONaX1000volume in ml

1000 mass of CH3COONa molarity of CH3C0ONa -X volume in ml molar mass of CH3COON2'

molarity of CH₃COONa=8.203 82.03aX1000100

8.203 1000 -x 100 molarity of CH3COONa 82.03a

           
molarity of CH₃COONa=1.0 M

molarity of CH3 COONa = 1.0 M

now we will calculate pH of Buffer by Using Henderson-Hasselbalch Equation            

pH=pKa+log[salt][Acid]

[salt pH 3 pКа +1og [Acid]

pKa of acetic acid = 4.76 (fixed value)

                                                            
pH=4.76+log[1][1]

[1] pH 4.76log [1]

on solving                                            pH=4.76

Answer 4

If we add 5 ml of 0.5 M NaOH the number of moles of CH₃COOH and CH₃COONa will change. so first we will calculate the number of moles of CH₃COOH and CH₃COONa after addition of NaOH

                        Number of moles = molarity X volume in liter

moles of CH3COOH in buffer

                        1 X .02 = 0.02 moles

moles of CH3COOH in buffer

                        1 X .02 = 0.02 moles

moles of NaOH

                        0.5X 0.005 = 0 .0025 moles

on adding NaOH it will react with acetic acid (decrease moles of acetic acid) and form sodium acetate (increase moles of sodium acetate) in 1:1 ratio. Hence moles of CH₃COOH and CH₃COONa after addition of NaOH is:

moles of CH3COOH

                             0.02 - 0.0025 = 0.0175 moles

                             0.02 + 0.0025 = 0.0225 moles

on putting in Henderson-Hasselbalch Equation                     

                     
pH=4.76+log[0.0225][0.01751]

0.0225 pH4.76+log To.01751]

                                                            pH= 4.76+0.0109

                                                            pH= 4.7709

Answer 2

For this pH of Buffer A and B is required

if we made our own data then at least it should be mentioned that whether we will take acidic buffer or basic buffer. composition of buffer will be same or different