A solution was prepared with 100 ml of 0.1 mol/L acetic acid and 100 ml of 0.1 mol/l sodium acetate. The density of acetic acid is 1.05 g/mL
Calculate the required volumes of each to prepare 25 mL of a buffer solution PH = 5.0. and calculate the PH. Data: Acetic acid pKa = 4.75
Wallace, your post is confusing. It start with A solution which tells me that the acetic acid and the sodium acetate are mixed BUT the question says to calculate the volumes of EACH which tells me they are two separate solutions. Please clear this up for me and tell me at the same time what you know about the Henderson-Hasselbalch equation.
To calculate the required volumes of acetic acid and sodium acetate to prepare a 25 mL buffer solution with pH 5.0, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
Where:
pH is the desired pH (5.0 in this case)
pKa is the dissociation constant of acetic acid (4.75 in this case)
[A-] is the concentration of the conjugate base (acetate ion, CH3COO-)
[HA] is the concentration of the acid (acetic acid, CH3COOH)
Let's calculate the concentrations of acetic acid and sodium acetate in the original solution.
The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. Given that 100 mL of 0.1 mol/L acetic acid and 100 mL of 0.1 mol/L sodium acetate were mixed, the number of moles of acetic acid and sodium acetate can be calculated as follows:
Moles of acetic acid (CH3COOH):
moles_CH3COOH = volume (L) x concentration (mol/L)
moles_CH3COOH = 0.1 mol/L x 0.1 L = 0.01 moles
Moles of sodium acetate (CH3COONa):
moles_CH3COONa = volume (L) x concentration (mol/L)
moles_CH3COONa = 0.1 mol/L x 0.1 L = 0.01 moles
Since the molar ratios of acetic acid and sodium acetate in the buffer solution are 1:1, we have equal moles of acetic acid and sodium acetate, which is 0.01 moles each.
To calculate the volume of each component needed to prepare a 25 mL buffer solution, we'll use the formula:
Volume (L) = Moles / Concentration (mol/L)
Volume of acetic acid (CH3COOH):
Volume_CH3COOH = moles_CH3COOH / concentration_CH3COOH
Volume_CH3COOH = 0.01 moles / 0.1 mol/L
Volume_CH3COOH = 0.1 L or 100 mL
Volume of sodium acetate (CH3COONa):
Volume_CH3COONa = moles_CH3COONa / concentration_CH3COONa
Volume_CH3COONa = 0.01 moles / 0.1 mol/L
Volume_CH3COONa = 0.1 L or 100 mL
Therefore, to prepare a 25 mL buffer solution with pH 5.0, you will need to mix 100 mL of acetic acid and 100 mL of sodium acetate.
To calculate the pH of the buffer solution, we'll use the Henderson-Hasselbalch equation mentioned earlier:
pH = pKa + log ([A-] / [HA])
Given that the concentration of acetic acid (HA) and sodium acetate (A-) is equal and will be 0.01 moles / 0.1 L = 0.1 mol/L, we can substitute these values into the equation:
pH = 4.75 + log (0.1 / 0.1)
pH = 4.75 + log (1)
pH = 4.75 + 0
pH = 4.75
Therefore, the pH of the buffer solution will be 4.75.