Can a buffer solution with pH = 4.70 be prepared using water, 6.0 M HC2H3O2 and 6.0 M NaOH? Justify your answer. Ka = 1.8 x 10-5 for HC2H3O2 , K = 1.0 x 10-14 .
The Henderson-Hasselbalch equation is
pH = pKa + log [(base)/(acid)]
pKa = 4.74 from Ka.
Generally it is easy to prepare a buffer within 1 pH of the pKa. You can work through the calculation if you wish; for example, if equal amounts of acid and base are chosen, then (B) = (A), the ratio is 1 and log of that term is 0 and pH = 4.74 which is very close to the desired pH.
After I went to bed I realized that part of my answer might be confusing. By (B)= (A), I am referring to base = acetate ion an acid = acetic acid. To GET equal amount of acetate ion and acetic acid, one reacts X amount of NaOH and twice that amount of acetic acid. That produces X amount of acetate ion, leaves an equal amount of acetic acid, and none of the NaOH you started with. I hope this helps clear things up.
To determine if a buffer solution with pH = 4.70 can be prepared using water, 6.0 M HC2H3O2, and 6.0 M NaOH, we need to first calculate the pH of the buffer that would be formed by mixing these two solutions.
A buffer solution is formed by combining a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, HC2H3O2 is a weak acid and its conjugate base is C2H3O2-.
To prepare a buffer with a specific pH, we need to calculate the ratio of the concentration of the weak acid to its conjugate base, known as the acid-to-base ratio (A/B) for the buffer. This ratio can be determined using the Henderson-Hasselbalch equation:
pH = pKa + log(A/B)
Where pKa is the negative logarithm of the acid dissociation constant (Ka) of the weak acid.
In this case, the given Ka for HC2H3O2 is 1.8 x 10^-5. To calculate the pKa, we take the negative logarithm of Ka:
pKa = -log(1.8 x 10^-5) ≈ 4.74
Now, let's substitute the given pH and pKa into the Henderson-Hasselbalch equation and solve for the acid-to-base ratio (A/B):
4.7 = 4.74 + log(A/B)
Subtracting 4.74 from both sides:
0.04 = log(A/B)
To convert this equation to exponential form and solve for A/B, we can rewrite it as:
10^0.04 = A/B
Which simplifies to:
1.096 = A/B
This means that for a buffer solution with pH = 4.70, the acid-to-base ratio (A/B) should be approximately 1.096.
Now, let's analyze the provided concentrations of HC2H3O2 and NaOH (6.0 M) and see if we can achieve the required A/B ratio.
Since the concentration of NaOH (a strong base) is given as 6.0 M, it will readily dissociate into Na+ and OH- ions. However, we need the weak base C2H3O2- to form the buffer with HC2H3O2.
The reaction between HC2H3O2 and NaOH would proceed as follows:
HC2H3O2 + NaOH → NaC2H3O2 + H2O
In this reaction, one molecule of NaOH reacts with one molecule of HC2H3O2 to form one molecule of NaC2H3O2 (the desired conjugate base) and one molecule of H2O.
Since the concentration of HC2H3O2 is also given as 6.0 M, we have the necessary concentration needed to form the buffer.
Now, let's calculate the concentration ratios of the weak acid (HC2H3O2) and its conjugate base (C2H3O2-) in the formed buffer:
A/B = [C2H3O2-] / [HC2H3O2]
Since the concentration of HC2H3O2 is 6.0 M, we can assume that the concentration of C2H3O2- will also be 6.0 M, as the reaction between HC2H3O2 and NaOH is balanced.
So, the concentration ratio A/B is:
A/B = 6.0 M / 6.0 M = 1.0
Comparing this calculated A/B ratio (1.0) with the required A/B ratio (1.096), we find that they do not match.
Therefore, a buffer solution with pH = 4.70 cannot be prepared using water, 6.0 M HC2H3O2, and 6.0 M NaOH. We would need to adjust the concentrations or use a different acid and conjugate base to achieve the desired pH.