Preparing Acetic Acid/Acetate Buffer pH 5.6, 250 mL

[Moe*]

You want to make an acetic acid/acetate buffer, pH 5.6, with a volume of 250 mL, and a final concentration of ([AcOH] + [Aco-]). You may only use acetic acid and sodium acetate( no strong acid or base). How many mL of glacial acetic acid would be needed( d=1.05g/mL)?

 

[chemisttree]

How might the Henderson-Hasselbach equation be used to solve this?

 

[Blueshift5]

To prepare a buffer solution with a pH of 5.6 using only acetic acid and sodium acetate, the Henderson-Hasselbalch equation can be used:

pH = pKa + log ([A-]/[HA])

Where pKa is the acid dissociation constant of acetic acid (4.76) and [A-] and [HA] are the concentrations of the acetate and acetic acid, respectively.

To achieve a pH of 5.6, we can set the equation up as follows:

5.6 = 4.76 + log ([A-]/[HA])

Solving for the ratio [A-]/[HA], we get a value of 0.398.

Next, we need to determine the amount of acetic acid and sodium acetate needed to achieve a final concentration of ([AcOH] + [Aco-]) in a 250 mL solution. Since we know the ratio of [A-]/[HA], we can set up the following equations:

[A-] = 0.398x
[HA] = x

Where x is the concentration of acetic acid and sodium acetate in mol/L.

Since we want a final volume of 250 mL, we can set up the following equation to find the total moles of acetic acid and sodium acetate needed:

250 mL = 0.398x + x

Solving for x, we get a value of 0.173 mol/L.

Next, we can convert this concentration into grams using the density of glacial acetic acid (1.05 g/mL):

0.173 mol/L x 60.05 g/mol = 10.38 g/L

To find the amount of acetic acid needed, we can use the following equation:

10.38 g/L x (250 mL/1000 mL) = 2.595 g or 2.47 mL

Therefore, approximately 2.47 mL of glacial acetic acid would be needed to make a 250 mL solution of acetic acid/acetate buffer with a pH of 5.6. The remaining volume (247.53 mL) would be made up of a 0.173 mol/L solution of sodium acetate.