How many milliliters of pure acetic acid (density: 1.09g/mL) must be added to 500mL of 0.1M NaOH solution to obtain such a buffer?
Iam not sure of the solution.
What buffer? See my solution below for pH 4.5 and different density of aceitic acid.
To determine how many milliliters of pure acetic acid are needed to make a buffer solution, we need to consider the chemical reaction between acetic acid (HC2H3O2) and sodium hydroxide (NaOH). The balanced chemical equation for the reaction is:
HC2H3O2 + NaOH -> H2O + NaC2H3O2
This reaction shows that one mole of acetic acid reacts with one mole of sodium hydroxide to produce one mole of water and one mole of sodium acetate.
First, let's calculate the number of moles of sodium hydroxide in the 500 mL of 0.1M NaOH solution:
Molarity (M) = moles/Liter
0.1M NaOH = x moles/0.5 L (500 mL = 0.5 L)
x = 0.1M NaOH * 0.5 L = 0.05 moles NaOH
Since the reaction is 1:1 between acetic acid and sodium hydroxide, we need the same number of moles of acetic acid to react with the sodium hydroxide to form a buffer.
Now let's calculate the mass of acetic acid needed using the molar mass of acetic acid (HC2H3O2):
Molar mass of HC2H3O2 = 2 * Atomic mass of C + 4 * Atomic mass of H + 2 * Atomic mass of O
= 2 * 12.01 g/mol + 4 * 1.01 g/mol + 2 * 16.00 g/mol
= 60.05 g/mol
The mass (g) of acetic acid needed is equal to the number of moles of acetic acid multiplied by its molar mass:
Mass = moles * molar mass
= 0.05 moles * 60.05 g/mol
= 3.003 g
Next, we can use the density of acetic acid (1.09 g/mL) to calculate the volume of pure acetic acid needed:
Volume = mass / density
= 3.003 g / 1.09 g/mL
≈ 2.75 mL
Therefore, approximately 2.75 milliliters of pure acetic acid need to be added to the 500 mL of 0.1M NaOH solution to obtain the desired buffer solution.