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Tanzania [10]
2 years ago
8

Write the formula of the conjugate acid of HCO₂⁻.

Chemistry
1 answer:
Alex2 years ago
7 0

Taking into account the Brønsted-Lowry acid-base theory, the conjugate acid of HCO₂⁻ is H₂CO₂.

<h3>Brønsted-Lowry acid-base </h3>

The Brønsted-Lowry acid-base theory (or the Brønsted-Lowry theory) identifies acids and bases based on whether the species accepts or donates protons or H⁺.

According to this theory, acids are proton donors while bases are proton acceptors. That is, an acid is a species that donates an H⁺ proton while a base is a chemical species that accepts an H⁺ proton from the acid.

So, reactions between acids and bases are H⁺ proton transfer reactions.

<h3>Conjugate base and conjugate acid</h3>

Then, a conjugate base is an ion or molecule resulting from the acid that loses the proton, while a conjugate acid is an ion or molecule resulting from the base that gains the proton:

acid + base ⇄ conjugate base + conjugate acid

<h3>Conjugate acid of HCO₂⁻</h3>

Like a conjugate acid is an ion or molecule resulting from the base that gains the proton, the conjugate acid of HCO₂⁻ is H₂CO₂.

Learn more about the Brønsted-Lowry acid-base theory:

<u>brainly.com/question/12916250?referrer=searchResults</u>

<u>brainly.com/question/1191429?referrer=searchResults</u>

<u>brainly.com/question/4000152?referrer=searchResults</u>

<u>brainly.com/question/12808135?referrer=searchResults</u>

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a scientist uses 68 grams of CaCo3 to prepare 1.5 liters of solution. what is the molarity of this solution?
victus00 [196]

Answer: 0.4533mol/L

Explanation:

Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol

68g of CaCO3 dissolves in 1.5L of solution.

Xg of CaCO3 will dissolve in 1L i.e

Xg of CaCO3 = 68/1.5 = 45.33g/L

Molarity = Mass conc.(g/L) / molar Mass

Molarity = 45.33/100 = 0.4533mol/L

7 0
3 years ago
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aleksandr82 [10.1K]

Answer:

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Explanation:

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The equilibrium concentrations of the reactants and products are [ HA ] = 0.260 M [HA]=0.260 M , [ H + ] = 2.00 × 10 − 4 M [H+]=
ValentinkaMS [17]

Answer:

pKa of the acid HA with given equilibrium concentrations is 6.8

Explanation:

The dissolution reaction is:

HA ⇔ H⁺ + A⁻

So at equilibrium, Ka is calculated as below

Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260

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Hence, by definition,

pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813

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