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You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol,MW=60.05 g/mol, pKa=4.76), pKa=4.76), 3.00 M NaOH,3.00 M NaOH, and water. Answer the questions regarding the preparation of the buffer. 1. How many grams of acetic acid will be needed to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

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jufsanabriasa

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀[tex]\frac{[A^{-}]}{[HA]}[/tex]

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀[tex]\frac{[A^{-}]}{[HA]}[/tex]

1,38 = [tex]\frac{[A^{-}]}{[HA]}[/tex] (1)

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] (2)

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid  ≡ 0,15 moles of acetic acid ×[tex]\frac{60,05 g}{1mol}[/tex] = 9,0 g of acetic acid

I hope it helps!

Lanuel

The quantity of acetic acid that is needed to prepare the 500 mL buffer is 9.0075 grams.

Given the following data:

  • Volume of acetate buffer = 500 mL to L = 0.5 L.
  • Molarity of acetate buffer = 0.300 M.
  • pH = 4.90.
  • MW = 60.05 g/mol.
  • pKa = 4.76.

How to calculate the mass of acetic acid.

First of all, we would write the equilibrium chemical reaction for acetate-acetic acid as follows:

                                [tex]CH_3COOH \rightleftharpoons CH_3COO^{-}+H^+[/tex]

Next, we would calculate HA by applying Henderson-Hasselbalch equation:

[tex]pH =pka+ log_{10} \frac{A^-}{HA}[/tex]

Where:

  • HA is acetic acid.
  • [tex]A^-[/tex]  is acetate ion.

Substituting the given parameters into the formula, we have;

[tex]4.90 =4.76+ log_{10} \frac{A^-}{HA}\\\\4.90 -4.76+ log_{10} \frac{A^-}{HA}\\\\\frac{A^-}{HA}=1.38\\\\A^- = 1.38[HA][/tex]

For the concentration of both acids, we have:

[tex][HA]+[A^-]=0.300M\\\\[HA]+1.38[HA]=0.300M\\\\2.38[HA]=0.300M\\\\HA = 0.126[/tex]

For acetate ion:

[tex]A^- = 1.38[HA] = 1.38 \times 0.126\\\\A^- =0.174[/tex]

At a volume of 0.5 liters, we have:

[tex]HA = 0.5 \times 0.126\\\\HA = 0.063 \;moles[/tex]

[tex]A^- = 0.5 \times 0.174\\\\A^- =0.087 \;moles[/tex]

By stoichiometry:

Total moles = [tex]0.063 + 0.087[/tex] = 0.15 moles.

[tex]Mass = number \;of \;moles \times molar\;mass\\\\Mass =0.15 \times 60.05[/tex]

Mass = 9.0075 grams.

Read more on moles here: brainly.com/question/3173452

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