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ELEN [110]
3 years ago
13

What is the Ka of a 0.0326 M solution of hydrofluoric acid with a pH of 3.94?

Chemistry
1 answer:
LenKa [72]3 years ago
8 0

Answer:

Kₐ = 4.06 × 10⁻⁷

Explanation:

Step 1. <em>Calculate [H₃O⁺] </em>

\text{H}_{3}\text{O}^{+} = 10^{\text{-pH}}\text{ mol/L}

pH = 3.94

\text{H}_{3}\text{O}^{+} = 10^{-3.94}} \text{ mol/L}

[H₃O⁺] = 1.15 × 10⁻⁴ mol·L⁻¹

Calculate K_{\text{a}}

                             HF + H₂O ⇌          H₃O⁺          +          F⁻

I/mol·L⁻¹:           0.0326                          0                         0

C/mol·L⁻¹: 0.0326-1.15 × 10⁻⁴        +1.15 × 10⁻⁴           +1.15 × 10⁻⁴

E/mol·L⁻¹:         0.0325                    1.15 × 10⁻⁴            1.15 × 10⁻⁴

So, at equilibrium,

[H₃O⁺] = [F⁻] = 1.15 × 10⁻⁴ mol·L⁻¹

[HF] = 0.326 – 1.15 × 10⁻⁴  mol·L⁻¹ = 0.0325 mol·L⁻¹

Kₐ = {[H₃O⁺][F⁻]}/[HF]

Kₐ = (1.15 × 10⁻⁴ × 1.15 × 10⁻⁴)/0.0325

Kₐ = 1.32 × 10⁻⁸/0.0325

Kₐ = 4.06 × 10⁻⁷

This is <em>NOT</em> a solution of HF (Kₐ = 7.2 × 10⁻⁴). It is more likely a solution of carbonic acid (H₂CO₃; Kₐ₁ = 4.27 × 10⁻⁷).

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1 gram of H2 will be produced from 12 grams of Mg.

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34g aluminum are combined with 39g chlorine gas which is the limiting reactant
vampirchik [111]

The limiting reactant is chlorine (Cl2).

<u>Explanation</u>:

Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.

        2 Al + 3 Cl2 ==> 2 AlCl3  represents the balanced equation.

Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g

                                                = 1.260 g moles of Al

Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g

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Dividing each reactant by it's coefficient in the balanced equation obtains:

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The reactant which produces a lesser amount of product is called as limiting reactant.

Here the Limiting reactant is Cl2.

6 0
3 years ago
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