Final answer:
To create a 100 mL of 0.2 M acetic acid buffer at pH 4.66, 0.785 mL of 17.6 N acetic acid and 0.8424 g of sodium acetate are required. The calculations are based on the mole ratio derived from the Henderson-Hasselbalch equation.
Step-by-step explanation:
To prepare 100 mL of a 0.2 M acetic acid and sodium acetate buffer at pH 4.66, we would need to use the Henderson-Hasselbalch equation to calculate the ratio of the concentrations of acetic acid to sodium acetate.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where [A-] is the concentration of the conjugate base (sodium acetate) and [HA] is the concentration of the acid (acetic acid).
Plugging in the given values we get:
4.66 = 4.76 + log([A-]/[HA])
This gives us a ratio [A-]/[HA] = 10-(4.76-4.66) = 0.7943
Since the total molarity of the buffer is 0.2 M, we can let x be the concentration of acetic acid and 0.2-x be the concentration of sodium acetate. Using the ratio:
x / (0.2 - x) = 0.7943
Solving for x, we find that x (concentration of acetic acid) is approximately 0.138 M and (0.2 - x) (concentration of sodium acetate) is approximately 0.062 M.
The normality of acetic acid is given as 17.6 N. Since normality is molarity multiplied by the number of protons an acid can donate, and acetic acid can donate only one proton, the molarity and normality in this case are the same. So the volume (V) of the 17.6 N acetic acid needed is calculated by:
V = n / N = (0.138 M * 0.1 L) / 17.6 N = 0.000785 L or 0.785 mL
To find the mass of sodium acetate, we use its molecular weight:
Mass of sodium acetate = moles * molecular weight = (0.062 M * 0.1 L) * 136 g/mol = 0.8424 g
Therefore, we would need 0.785 mL of 17.6 N acetic acid and 0.8424 g of sodium acetate to prepare 100 mL of the desired buffer solution.