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Juli2301 [7.4K]
3 years ago
14

The ionization constant (Ka) of HF is 6.7 X 10 ^-4. Which of the following is true in a 0.1 M solution of this acid?

Chemistry
2 answers:
spin [16.1K]3 years ago
8 0
The ionization equation is:

HF ⇄ H(+) + F(-)

The ionization constant is Ka = [H(+)] * [H(-)] / [HF]

=> [H(+)] * [F(-)] = Ka * [HF]

Given that Ka < 1

[H(+)] * [F(-)] < [HF]

Which is [HF] >  [H(+)] * [F(-)] the option a. fo the list of choices.
n200080 [17]3 years ago
6 0

The correct answer is [HF] is less than [H+][F-]

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1 year ago
A 33.153 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis ap
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Answer:

The empirical formula of the compound = C_4H_8S_1O_1

Explanation:

Mass of carbon dioxide gas = 59.060 mg = 0.059060 g

1 mg = 0.001 g

Moles of carbon dioxide = \frac{0.059060 g}{44 g/mol}=0.0013 mol

Moles of carbon in 0.0013 moles of carbon dioxide gas = 1 × 0.0013 mol = 0.0013 mol

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Mass of water = 24.176 mg = 0.024176

Moles of water = \frac{0.024176 g}{18 g/mol}=0.0013 mol

Moles of hydrogen in 0.0013 moles of water = 2 × 0.0013 mol = 0.0026 mol

Mass of 0.0013 moles of hydrogen= 1 g/mol\times 0.0013 mol=0.0013 g

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Hydrogen: \frac{0.0026 mol}{0.00032 mol}=8

Sulfur : \frac{0.00032 mol}{0.00032 mol}=1

Oxygen : \frac{0.00038 mol}{0.00032 mol}=1

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