Thanks for answering the first one :)
Here is another
Calculate the amount of sodium acetate and acetic acid necessary to make 100mL of a 0.500 M acetate buffer that is pH 5.2.
I believe I have the answer but I want to make sure I did the work correctly.
I have the pka from previous problems 4.75
To calculate the amount of sodium acetate and acetic acid needed to prepare a 0.500 M acetate buffer with a target pH of 5.2, you would need to consider the Henderson-Hasselbalch equation and the relationship between pH, pKa, and the concentrations of the conjugate acid-base pair (acetic acid/acetate).
The Henderson-Hasselbalch equation is: pH = pKa + log([A-]/[HA])
In this case, acetic acid (HA) is the weak acid and acetate (A-) is its conjugate base. The pKa of acetic acid is a constant value that can be found using available references. For acetic acid, the pKa is 4.76.
To calculate the concentrations of the acid and salt required, you need to determine the ratio of [A-]/[HA] that corresponds to the desired pH value.
1. Start by substituting the given values into the Henderson-Hasselbalch equation: 5.2 = 4.76 + log([A-]/[HA])
2. Rearrange the equation to solve for [A-]/[HA]: [A-]/[HA] = 10^(pH - pKa)
3. Since a 0.500 M concentration is desired, you can assign one of the concentrations and solve for the other. Let's assign [A-] = 0.500 M.
4. Substitute the assigned value and calculate [HA]: [A-]/[HA] = 10^(5.2 - 4.76)
[A-]/[HA] = 10^0.44
[A-]/[HA] = 1.359
5. Rearrange the ratio equation to solve for [HA]: [HA] = [A-]/1.359
[HA] = 0.500/1.359
[HA] ≈ 0.368 M
So, to make 100 mL of a 0.500 M acetate buffer at pH 5.2, you would need approximately 0.368 moles (36.8 grams) of acetic acid (HA) and 0.500 moles (82.5 grams) of sodium acetate (A-). Keep in mind that these values assume ideal behavior and do not account for the actual solubility limits of the compounds in the solvent being used.