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Kaylis [27]
3 years ago
5

In class, it was mentioned that at 25 °C, the pH of pure water is 7.00 becuase Kw = 1.0 × 10−14. At other temperatures, the pH v

alue of pure water will be different than 7.00 becuase the value of Kw changes. Suppose that at a temperature different than 25 °C, the pH of pure water was measured and found to be 7.27. What must the value of Kw be equal to at this second temperature?
Chemistry
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

Kw = 2.88 × 10⁻¹⁵

Explanation:

Let's consider the dissociation of water.

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

The equilibrium constant Kw is:

Kw = [H⁺].[OH⁻]

If pH = 7.27, we can find [H⁺]:

pH = -log [H⁺]

H⁺ = anti log (-pH) = anti log (-7.27) = 5.37 × 10⁻⁸ M

According to the balanced equation, 1 mole of H⁺ is produced per mole of OH⁻. So, [H⁺] = [OH⁻] = 5.37 × 10⁻⁸ M

Then,

Kw = [H⁺].[OH⁻]= (5.37 × 10⁻⁸)² = 2.88 × 10⁻¹⁵

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The density of water at 30.0 °C is 0.9956 g/mL. If the specific gravity of acetic acid is 1.040 at 30.0 °C, what is the density
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Answer:

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Explanation:

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A buffer with a pH of 3.98 contains 0.23 M of sodium benzoate and 0.38 M of benzoic acid. What is the concentration of [H3O+] in
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Answer:

New pH = 3.84

Explanation:

First of all we may think that if the buffer has pH 3.98 and we're adding H⁺, pH's buffer will be lower, as the [H⁺] is been increased.

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0.38 M . 1.3L = 0.494 moles of HBz

We add 0.058 of HCl, which is the same as 0.058 moles of H⁻

HCl →  H⁺  +  Cl⁻

As we add the moles of protons, these are going to react to the Bz⁻

In the buffer system we have these dissociations:

NaBz  →  Na⁺  +  Bz⁻

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So, as we add protons, we have a new equilibrium:

          Bz⁻      +      H⁺       ⇄   HBz    

In    0.299         0.058          0.494

Eq   0.241               -              0.552

Protons are substracted to benzoate, so the [HBz] is now higher than before. We calculate the new pH, with the Henderson Hasselbach equation

pH = pKa + log (Bz⁻/HBz)

pH = 4.20 + log (0.241 / 0.552) → 3.84

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2 years ago
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