To make a solution with a pH = 4.40 a student used the following procedure :
a certain amount of sodium acetate along with 0.280 moles of acetic acid is added to enough water to make a solution of 1.00 L
How many grams of sodium acetate were added ?
Ka = 1.80 x 10 -5
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid)
To find out how many grams of sodium acetate were added, we need to follow these steps:
Step 1: Determine the balanced chemical equation for the reaction between sodium acetate (CH3COONa) and acetic acid (CH3COOH).
The balanced equation is:
CH3COOH + CH3COONa -> CH3COONa + H2O
Step 2: Find the molar mass of sodium acetate (CH3COONa).
Using the periodic table, we find the molar mass of each component:
- Carbon (C) = 12.01 g/mol
- Hydrogen (H) = 1.008 g/mol
- Oxygen (O) = 16.00 g/mol
- Sodium (Na) = 22.99 g/mol
So, the molar mass of sodium acetate (CH3COONa) = (12.01 * 2) + (1.008 * 3) + 16.00 + 22.99 = 82.03 g/mol.
Step 3: Calculate the number of moles of acetic acid (CH3COOH) used.
We are given that 0.280 moles of acetic acid (CH3COOH) is used in the reaction.
Step 4: Use the balanced equation to determine the stoichiometry between sodium acetate (CH3COONa) and acetic acid (CH3COOH).
From the balanced equation, we know that one mole of acetic acid (CH3COOH) reacts with one mole of sodium acetate (CH3COONa).
Step 5: Calculate the moles of sodium acetate (CH3COONa) required.
Since the stoichiometry between sodium acetate and acetic acid is 1:1, the moles of sodium acetate required will also be 0.280 moles.
Step 6: Calculate the mass of sodium acetate (CH3COONa) required.
Using the equation: mass = moles × molar mass, we can calculate the mass of sodium acetate required:
mass = 0.280 moles × 82.03 g/mol = 22.89 grams.
Therefore, 22.89 grams of sodium acetate were added to make the solution with a pH of 4.40.