Chemistry, 01.07.2019 12:30 jself

What is the ka of a 0.0326 m solution of hydrofluoric acid with a ph of 3.94?

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Answer from: DEEWILL5100

Kₐ = 4.06 × 10⁻⁷

Explanation:

Step 1. Calculate [H₃O⁺]

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

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

Answer from: Quest

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

explanation:

Answer from: Quest

1. and 2. question: in which equation? or i don't understand you, 3. a), 4. c), 5. c), 6. b), 7. d), 8. d) if small number is underlined(behind symbol) , b) if big number is underlined(in front of symbol)9. what is underlined there? 10. c)

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