Question

2. Calculate the volume of 4.2 M acetic acid needed to prepare 100 mL of a 0.10 M acetic acid (CH3COOH) solution. Calculate the pH of the resulting solution.

3. Calculate the mass of sodium acetate (CH3COONa) required to prepare 35 mL of a 0.15 M sodium acetate solution. Calculate the pH of the resulting solution.

Answer 1

To prepare a 0.10 M acetic acid solution, you will need approximately 2.38 mL of 4.2 M acetic acid. The pH of the resulting solution will be approximately 4.75.

Step-by-step explanation:

To calculate the volume of 4.2 M acetic acid needed to prepare 100 mL of a 0.10 M acetic acid solution, we can use the formula:

Volume of 4.2 M acetic acid = (0.10 M)(100 mL) / 4.2 M

This gives us a volume of approximately 2.38 mL of 4.2 M acetic acid needed.

To calculate the pH of the resulting solution, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([conjugate base]/[acid])

In this case, the pKa for acetic acid is 4.75. Plugging in the values, we get:

pH = 4.75 + log(0.10/0.10)

After simplification, the pH of the resulting solution is approximately 4.75.

Answer 2

2) The pH of the solution is 1

3) The mass of the sodium acetate is 0.43 g. The pH of the solution is 11.

What is dilution in chemistry?

Using;

C1V1 = C2V2

4.2 * V = 100 * 0.10

V = 100 * 0.10/4.2

V = 2.4 L

The pH of the solution is -log(0.1)

= 1

3) Moles = Concentration * volume

= 0.15 M * 35/1000 L

= 0.00525 moles

Mass of the sodium acetate = 0.00525 moles * 82 g/mol

= 0.43 g

The dissociation equation is;

`CH_3COO^- + H_2O ---- > CH_3COOH + OH^-`

The Kb of acetic acid is
`5.56*10^{-10`

`5.56*10^{-10` =
`x^2`/0.15 - x

`5.56*10^{-10` (0.15 - x) =
`x^2`

8.34 *
`10^{-11` -
`5.56*10^{-10`x =
`x^2`

`x^2` +
`5.56*10^{-10`x - 8.34 *
`10^{-11` = 0

x = 0.0000091 M

pOH = -log(0.0000091)

pOH = 5

pH = 14 - 5

pH = 11