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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:

6.25\times 10^{-6} is the value of the equilibrium constant at this temperature.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures  of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as K_{p}

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Answer:

3.59x10^21 molecules

Explanation:

1mole of a substance contains 6.02x10^23 molecules.

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1mole (i.e 114g) oh C8H18 contains 6.02x10^23 molecules.

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If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
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Answer:

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The corrected balanced reaction equation is:

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(2.0 g) / (2.29 g) x 100% = 87%

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