Question

A researcher prepares a buffer of acetic acid and sodium acetate with a pH of 5.0.5.0. The total concentration of both components in the buffer is 250 mM,250 mM, and acetic acid has a pKapKa of 4.75.4.75. What are the concentrations of acetic acid and sodium acetate in the buffer? concentration of acetic acid: mM concentration of sodium acetate: mM How many moles of acetic acid and sodium acetate are in 2 L2 L of the buffer? amount of acetic acid: mol amount of sodium acetate: mol How many grams of acetic acid and sodium acetate are in 2 L2 L of the buffer? The molar mass of acetic acid is 60.05 g/mol;60.05 g/mol; the molar mass of sodium acetate is 82.03 g/mol.82.03 g/mol. mass of acetic acid: g mass of sodium acetate:

Answer 1

Answer:

The concentrations are 89.98 mM for Acetic acid and 160.02 mM for the Sodium acetate

The 2L solution  of buffer contain 0.17996 moles of Acetic acid and 0.32004 molesof Sodium acetate.

The 2 L solution of buffer has 10.8 g of acetic acid and 26.3 g of sodium acetate-

Explanation:

Part 1

Hendersson Haselbalch equation relates the pH, pKa and the concentrations of acid and conjugated base:

$pH = pKa + log(\frac{base}{acid} )$

The total concentration of the buffer is 250mM, which means that the addition of the concentration of acid and base must be equal to 250 mM:

$[Acetic-acid] + [Sodium-acetate] = 250 mM$

Rearranging the equation we can get:

$[Sodium-acetate]=250mM - [Acetic-acid]$

Plugging the values of pH, pKa and replacing the concentration of sodium acetate (base) in the Hendersson Haselbalch equation we get:

$5 = 4.75 + log\frac{(250mM-[Acetic -acid]}{[Acetic-acid]}$

Then we solve to find the concentration of acetic acid in the buffer:

$5-4.75=log\frac{250mM-[Acetic-acid]}{[Acetic-acid]}$

$0.25=log\frac{250mM-[Acetic-acid]}{[Acetic-acid]}$

$10^{0.25} =\frac{250mM-[Acetic-acid]}{[Acetic-acid]}$

$1.7783 * [Acetic-acid] = 250mM-[Acetic-acid]$

$1.7783*[Acetic-acid] + [Acetic-acid] = 250 mM$

$2.7783*[Acetic-acid] = 250 mM$

$[Acetic-acid] =\frac{250mM}{2.7783}$

$[Acetic-acid]=89.98 mM$

To obtain the sodium acetate concentration we do:

$250 mM -89.98 mM = 160.02 mM$

Part 2

Using the molarity formula:

$Molarity = \frac{moles}{Volume (L)}$

we replace the corresponding molarities to find the mmoles:

For acetic acid:

$89.98 mM = \frac{mmoles}{2L}$

$89.98 mM* 2 L = 179.96 mmoles$  or 0.17996 moles

For Sodium acetate:

$160.02 mM = \frac{mmoles}{2L}$

$160.02 mM* 2 L = 320.04 mmoles$ or 0.32004 moles

Part 3

To convert the moles into masswe use the molar mass of the compound.

For acetic acid:

$0.17996 moles-acetic-acid * \frac{60.05 g-acetic-acid}{1 mol-acetic-acid} = 10.8 g-acetic-acid$

For sodium acetate:

$0.32004moles-sodium-acetate *\frac{82.03g-sodium-acetate}{1mol-sodium-acetate} = 26.3g-sodium-acetate$