Suppose you made a buffer solution that was 0.050M in both HC_2H_3O_2 and NaC_2H

Question

Suppose you made a buffer solution that was 0.050M in both HC_2H_3O_2 and NaC_2H_3O_2. Would the pH changes resulting from the addition of 0.1 M HCl solution to this buffer solution be the same as those you observed in this experiment? Briefly explain. Calculate the pH of a buffer solution that is 0.40M in both HC_2H_3O_2. Calculate the pH of a buffer solution that is 0.10M in HC_2H_3O_2 and NaC_2H_3O_2. Assume you have 1 L of each of the buffer solutions described in Parts I and II. To which buffer solution could you add the larger volume of 0.10M NaOH solution without causing a pH change of more than one or two pH units? Briefly explain. Calculate K_a for the weak acid HA, given that a buffer solution that is 0.30M in HA and 0.30M in NaA has a pH of 5.03.

Explanation / Answer

1)

On addition of of 0.1M HCl soution to the buffer solution the resulting pH will vary because the Conc. of H+ will increases so that will change the value of the conc. in the equattion.

2)

you must know that this is an example of a buffer solution.

Buffer solutions are solutions of weak acids or weak base with its salt.

For your question, the weak acid is acetic acid, and the salt is sodium acetate.

For buffer solutions, the Henderson-Hasselbach equation is used. It is defined as:

pH = pKa + log (salt concentration/acid concentration)

In your example, Ka is given. Thus, you must get the pKa.
pKa = -log Ka = - log 1.8 x 10^-5 = 4.74

Then you compute the pH,

pH = 4.74 + log (0.40/0.40) = 4.74

the answer is 4.74.

3)

For the 2nd question the value of salt conc. = 0.40 M

Acid conc. = 0.1M

pH = 4.74 + log (0.40/0.10) = 5.34

3)

The 1st solution will not change that much of pH

as log(5/4) < log(5)

4) 'According to
the Henderson-Hasselbach equation is used. It is defined as:

pH = pKa + log (salt concentration/acid concentration)

SO 5.03 = pKa + log (0.3 / 0.3)

=> 5.03 = -log Ka

=> Ka = 9.3 * 10^-6

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