Acetic acid has a pK of 4.8. How many mL of 0.1M acetic acid and 0.1M of sodium acetate are required to prepare 1 liter of 0.1M buffer solution having a pH of 5.8?
To calculate the required amounts of acetic acid and sodium acetate to prepare a buffer solution with a pH of 5.8, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, pH is the desired pH of the buffer solution, pKa is the acid dissociation constant of the acid (acetic acid in this case), [A-] is the concentration of the conjugate base (sodium acetate), and [HA] is the concentration of the acid (acetic acid).
Given information:
- pKa of acetic acid = 4.8
- pH of the buffer solution = 5.8
Let's start by using the Henderson-Hasselbalch equation to find the ratio of [A-]/[HA]:
5.8 = 4.8 + log([A-]/[HA])
1 = log([A-]/[HA])
Now, we need to convert the logarithmic equation into an exponential form:
10^1 = [A-]/[HA]
10 = [A-]/[HA]
Since we want to prepare a 0.1M buffer solution (both [A-] and [HA] have the same concentration), let's assume the concentration of sodium acetate ([A-]) and acetic acid ([HA]) are both x M.
For sodium acetate ([A-]):
[A-] = x M
For acetic acid ([HA]):
[HA] = x M
Now let's substitute these values into the equation:
10 = [A-]/[HA]
10 = (x M)/(x M)
10 = 1
This equation tells us that the ratio of [A-] to [HA] should be 1:1. Therefore, we need equal amounts of acetic acid and sodium acetate to prepare the buffer solution.
Since we need to prepare a 0.1M buffer solution with a total volume of 1 L, the amount of acetic acid and sodium acetate needed will be:
Volume of acetic acid = 0.1 M * 1 L = 0.1 moles
Volume of sodium acetate = 0.1 M * 1 L = 0.1 moles
Thus, you would need 0.1 moles (or 100 mL) of 0.1M acetic acid and 0.1 moles (or 100 mL) of 0.1M sodium acetate to prepare 1 liter of the 0.1M buffer solution with a pH of 5.8.