Acetic acid has a pKa of 4.8. How many milliliters of 0.2 M acetic acid and 0.1 M sodium acetate are required to prepare 1 liter of 0.1 M buffer solution having a pH of 4.8? Question 7 options: 500 ml acetic acid and 250 ml sodium acetate, then 250 ml water 250 ml acetic acid and 250 ml sodium acetate, then 500 ml water 250 ml acetic acid and 500 ml sodium acetate, then 250 ml water 500 ml acetic acid and 500 ml sodium acetate
Question
Acetic acid has a pKa of 4.8. How many milliliters of 0.2 M acetic acid and 0.1 M sodium acetate are required to prepare 1 liter of 0.1 M buffer solution having a pH of 4.8? Question 7 options:
500 ml acetic acid and 250 ml sodium acetate, then 250 ml water
250 ml acetic acid and 250 ml sodium acetate, then 500 ml water
250 ml acetic acid and 500 ml sodium acetate, then 250 ml water
500 ml acetic acid and 500 ml sodium acetate
Solution
To solve this problem, we need to use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the base (sodium acetate) and [HA] is the concentration of the acid (acetic acid).
Given that the pH is equal to the pKa, the log term in the equation becomes 0, meaning that the concentrations of the acid and the base must be equal. Therefore, we need equal amounts of acetic acid and sodium acetate.
Since the final buffer solution needs to be 0.1 M and we need to prepare 1 liter of it, we need 0.1 moles of each the acid and the base.
To get 0.1 moles of acetic acid from a 0.2 M solution, we need 0.1/0.2 = 0.5 liters or 500 ml of the acetic acid solution.
To get 0.1 moles of sodium acetate from a 0.1 M solution, we need 0.1/0.1 = 1 liter or 1000 ml of the sodium acetate solution.
However, adding 500 ml of acetic acid and 1000 ml of sodium acetate would result in a total volume of 1500 ml, which is more than the 1 liter we want to prepare.
To solve this, we can reduce the volume of the sodium acetate solution by half, to 500 ml. This will give us 0.05 moles of sodium acetate, which is half of what we need. To make up for the other half, we can add 500 ml of the acetic acid solution, which will also give us 0.05 moles of acetate ions (since acetic acid is a weak acid and will partially ionize to give acetate ions).
So, the correct answer is 500 ml acetic acid and 500 ml sodium acetate.
Similar Questions
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