a student needed 1000 ml of a 0.10 buffer of ph 4.0. calculate the amount of sodium acetate and acidic acid needed for this solution
I assume you meant acetic acid (and not acidic acid). Also, I assume you mean to add M so that the sentence reads .....1000 mL of a 0.10 M buffer (I have added M).
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/{acid)]
pKa = 4.74 for acetic acid.
You want pH = 4.00. Substitute 4.00 and 4.74 appropriately in the equation and calculate (base)/(acid). I found B/A = 0.182 or B = 0.182*A.
So we take 1000 mL of 0.1 M acetic acid (M x L = 0.1 mole). Use the ratio you calculated to determine moles base and go from there to grams sodium acetate. Post your work if you get stuck. I think the answer is about 1 or 2 grams NaC2H3O2.
To calculate the amount of sodium acetate and acetic acid needed for the buffer solution, we need to consider the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentration of the acid and its conjugate base.
Henderson-Hasselbalch Equation:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH of the buffer solution (pH = 4.0)
- pKa is the dissociation constant of the acid (acetic acid) and represents the pH at which half the acid is dissociated into its conjugate base
- [A-] is the concentration of the conjugate base (sodium acetate)
- [HA] is the concentration of the acid (acetic acid)
To calculate the amounts, we need to follow these steps:
Step 1: Calculate the moles of sodium acetate and acetic acid needed.
Moles = Volume (in liters) x Molarity
Given:
- Volume = 1000 mL = 1 L
- Buffer pH = 4.0 (pKa for acetic acid = 4.74)
Using the Henderson-Hasselbalch equation, we can rearrange it to solve for [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
[A-]/[HA] = 10^(4.0 - 4.74)
[A-]/[HA] = 10^(-0.74)
[A-]/[HA] = 0.1869
Let's assume x moles of sodium acetate and acetic acid are needed. Then, the moles of the conjugate base [A-] would be 0.1869x, and the moles of the acid [HA] would be x.
Step 2: Calculate the moles of each component.
Moles of sodium acetate (x) = 0.1869x
Moles of acetic acid (x) = x
Step 3: Convert moles to grams using molar mass.
Molar mass of sodium acetate (CH3COONa) = 82.03 g/mol
Molar mass of acetic acid (CH3COOH) = 60.05 g/mol
Mass of sodium acetate = Moles of sodium acetate x Molar mass of sodium acetate
Mass of acetic acid = Moles of acetic acid x Molar mass of acetic acid
Now you can calculate the mass of sodium acetate and acetic acid needed for the buffer solution.