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DochEvi [55]
3 years ago
10

Buffer preparation. You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acid plus acet

ate concentration of 250 mM and a pH of 5.0. What concentrations of acetic acid and sodium acetate should you use? Assuming you wish to make 2 liters of this buffer, how many moles of acetic acid and sodium acetate will you need? How many grams of each will you need (molecular weights: acetic acid 60. 05 g mol 1, sodium acetate, 82. 03 g mol 1)?
Chemistry
1 answer:
Hoochie [10]3 years ago
8 0

Answer:

0.182 moles of acetic acid are needed, this means 10.93 g.

0.318 moles of sodium acetate are needed, this means 26.08 g.

Explanation:

The Henderson–Hasselbalch (<em>H-H</em>) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:

pH = pka + log\frac{[A^{-} ]}{[HA]}

In this case, [A⁻] is the concentration of sodium acetate, and [HA] is the concentration of acetic acid. The pka is a value that can be looked up in literature: 4.76.

From the problem we know that

[A⁻] + [HA] = 250 mM = 0.250 M     eq. 1

We use the <em>H-H</em> equation, using the data we know, to describe [A⁻] in terms of [HA]:

5.0 = 4.76 + log\frac{[A^{-} ]}{[HA]}

0.24=log\frac{[A^{-} ]}{[HA]}\\\\10^{0.24}=\frac{[A^{-} ]}{[HA]}\\ 1.74 [HA] = [A^{-}]        eq.2

Now we replace the value of [A⁻] in eq. 1, to calculate [HA]:

1.74 [HA] + [HA] = 0.250 M

[HA] = 0.091 M

Then we calculate [A⁻]:

[A⁻] + 0.091 M = 0.250 M

[A⁻] = 0.159 M

Using the volume, we can calculate the moles of each substance:

  • moles of acetic acid = 0.091 M * 2 L = 0.182 moles
  • moles of sodium acetate = 0.159 M * 2 L = 0.318 moles

Using the molecular weight, we can calculate the grams of each substance:

  • grams of acetic acid = 0.182 mol * 60.05 g/mol = 10.93 g
  • grams of sodium acetate =  0.318 mol * 82.03 g/mol = 26.08 g

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Answer:

pH = 9.48

Explanation:

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and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.

Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:

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mol NH₃ = 0.139 L x 0.39 M = 0.054 mol

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Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)

pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52

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How many grams of NaHCO3 are needed to prepare 250 mL of 0.50 M<br> NaHCO3?
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Answer:

10.5g

Explanation:

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Molarity = mole /Volume

Mole = Molarity x Volume

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Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol

Number of mole of NaHCO3 = 0.125 mole

Mass of NaHCO3 =?

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Mass of NaHCO3 = 0.125 x 84

Mass of NaHCO3 = 10.5g

Therefore, 10.5g of NaHCO3 is needed.

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