acetic acid has a pka of 4.8 how many ml of 0.1M acetic acid and 0.1M sodium acetate are required to prepare 1L of 0.1M buffer with a PH of 5.8? solve
To solve this question, we need to understand the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa of the acid component and the ratio of its concentrations to its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
In this equation, pH is the desired pH of the buffer, pKa is the acid dissociation constant, [A-] is the concentration of the acetate (conjugate base), and [HA] is the concentration of acetic acid (acid component). We need to rearrange this equation to solve for the ratio of [A-]/[HA].
1. Rearrange the equation to isolate the ratio [A-]/[HA]:
pH - pKa = log([A-]/[HA])
2. Take the antilog of both sides:
10^(pH - pKa) = [A-]/[HA]
3. Substitute the values given:
10^(5.8 - 4.8) = [A-]/[HA]
10^1 = [A-]/[HA]
[A-] = [HA]
This means that the ratio of the concentration of the conjugate base to the acid component should be 1:1.
Now, let's calculate the quantities of acetic acid and sodium acetate required to prepare the buffer:
1. Calculate the moles of acetic acid required to prepare 1L of 0.1M buffer:
Moles of acetic acid = concentration × volume
Moles of acetic acid = 0.1 mol/L × 1 L = 0.1 mol
2. Calculate the moles of sodium acetate required to prepare 1L of 0.1M buffer:
Moles of sodium acetate = concentration × volume
Moles of sodium acetate = 0.1 mol/L × 1 L = 0.1 mol
3. Calculate the required volumes of 0.1M acetic acid and 0.1M sodium acetate:
Volume of 0.1M acetic acid = (moles of acetic acid) / (concentration of acetic acid)
Volume of 0.1M acetic acid = 0.1 mol / 0.1 mol/L = 1 L
Volume of 0.1M sodium acetate = (moles of sodium acetate) / (concentration of sodium acetate)
Volume of 0.1M sodium acetate = 0.1 mol / 0.1 mol/L = 1 L
Therefore, you would need 1L of 0.1M acetic acid and 1L of 0.1M sodium acetate to prepare a 1L buffer solution with a pH of 5.8.