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andreev551 [17]
4 years ago
11

When a hydrochloric acid solution is combined with a potassium hydroxide solution, an acid-base reaction occurs Write a balanced

molecular equation for this reaction. Express your answer as a chemical equation. Identify all of the phases in your answer..
Chemistry
1 answer:
Svetach [21]4 years ago
7 0

Answer:

HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)

Explanation:

Hydrochloric acid is an acid because it releases H⁺ in an aqueous solution.

Potassium hydroxide is a base because it releases OH⁻ in an aqueous solution.

When an acid reacts with a base they form a salt and water. This is a neutralization reaction. The neutralization reaction between hydrochloric acid and potassium hydroxide is:

HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)

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Explanation:

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In an atom the number of electron is what as the number of protons???​
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How cations can become neutral atoms?​
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Be sure to answer all parts. A 0.365−mol sample of HX is dissolved in enough H2O to form 835.0 mL of solution. If the pH of the
Marta_Voda [28]

Answer:

The Ka is 9.11 *10^-8

Explanation:

<u>Step 1: </u>Data given

Moles of HX = 0.365

Volume of the solution = 835.0 mL = 0.835 L

pH of the solution = 3.70

<u>Step 2:</u> Calculate molarity of HX

Molarity HX = moles HX / volume solution

Molarity HX = 0.365 mol / 0.835 L

Molarity HX = 0.437 M

<u />

<u>Step 3:</u> ICE-chart

[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4

Initial concentration of HX = 0.437 M

Initial concentration of X- and H3O+ = 0M

Since the mole ratio is 1:1; there will react x M

The concentration at the equilibrium is:

[HX] = (0.437 - x)M

[X-] = x M

[H3O+] = 1.995*10^-4 M

Since 0+x = 1.995*10^-4   ⇒ x=1.995*10^-4

[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M

[X-] = x = 1.995*10^-4 M

<u>Step 4: </u>Calculate Ka

Ka = [X-]*[H3O+] / [HX]

Ka = ((1.995*10^-4)²)/ 0.437

Ka = 9.11 *10^-8

The Ka is 9.11 *10^-8

4 0
3 years ago
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