Final answer:
To prepare 3 L of a 0.2 M acetate buffer at pH 5.0, dissolve 74.12 g of Sodium Acetate and add 54.5 mL of 1 N Acetic Acid after calculating the ratio from the Henderson-Hasselbalch equation.
Step-by-step explanation:
The question asks to prepare a 0.2 M acetate buffer with a pH of 5.0 using Sodium Acetate and 1 N Acetic Acid, with a given pKa of 4.76. To prepare this buffer, the Henderson-Hasselbalch equation must be applied: pH = pKa + log([A-]/[HA]). From the pH (5.0) and pKa (4.76), calculate the ratio of acetate anion ([A-]) to acetic acid ([HA]). For a pH of 5.0, the ratio should roughly be about 10:1 of acetate anion to acetic acid due to the pH being one unit above the pKa. We can start by assuming we'll have 0.2 mol/L of total acetate (both the sodium acetate and acetic acid). Given the desired volume of 3 L, the total moles of acetate needed would be 0.2 mol/L * 3 L = 0.6 moles. With the 10:1 ratio, we calculate 0.545 mol for sodium acetate and 0.0545 mol for acetic acid.
Now, calculate the masses and volumes required: To find the mass of Sodium Acetate needed, multiply the moles (0.545 mol) by the molecular weight (136 g/mol), obtaining 74.12 g of Sodium Acetate. For the volume of 1 N Acetic Acid needed, use its molarity (N) and the moles required (0.0545 mol) which equals 54.5 mL. Therefore, to make 3 L of a 0.2 M buffer at pH 5.0, you need to dissolve 74.12 g of Sodium Acetate and add 54.5 mL of 1 N Acetic Acid.