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Karolina [17]
3 years ago
6

How does the value of Ka and Kb compare with the concentrations of hydronium ion and hydroxide ion?

Chemistry
2 answers:
Andrews [41]3 years ago
8 0
Assuming that we are given the values of Ka and Kb, these can be rewritten and represented by the concentration of the Hydronium ion and Hydroxide Ion:

Hydronium ion: [H3O+]
Hydroxide ion: [OH-]

Ka is the equilibrium constant of an acid which dissociates to a hydronium ion and its cation.
Kb is the equilibrium constant of a base which dissociates to a hydroxide ion and its anion. 

So,

Ka x Kb = [H3O+][OH-]
Nadusha1986 [10]3 years ago
8 0

Answer:

The correct answer is A, Ka × Kb = [H₃0⁺][OH⁻]

I hope this helps, have a good day and good luck! :)

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Calculate the pH of each of the following aqueous solutions. (Enter your answers to two decimal places.) (a) 10.0 mL deionized w
Anna007 [38]

Answer:

a. pH = 7.0

b. pH = 12.52

c. pH = 12.70

d. pH = 12.78

Explanation:

a. Deionized water has the [H⁺] of pure water = 1x10⁻⁷ (Kw = 1x10⁻¹⁴ = [H⁺][OH⁻] - [H⁺] = [OH⁻ -)

pH = -log[H⁺] = 7

b. Moles NaOH = 5x10⁻³L * (0.10mol / L) = 5x10⁻⁴moles OH⁻ / 0.015L = 0.0333M = [OH⁻]

<em>-Total volume = 10mL+5mL = 15mL = 0.015L</em>

pOH = -log[OH⁻] = 1.48

pH = 14-pOH

pH = 12.52

c. Moles NaOH = 0.010L * (0.10mol / L) = 1x10⁻³moles OH⁻ / 0.020L = 0.0500M = [OH⁻]

<em>-Total volume = 10mL+10mL = 20mL = 0.020L</em>

pOH = -log[OH⁻] = 1.30

pH = 14-pOH

pH = 12.70

d. Moles NaOH = 0.015L * (0.10mol / L) = 1.5x10⁻³moles OH⁻ / 0.025L = 0.060M = [OH⁻]

<em>-Total volume = 10mL+15mL = 25mL = 0.025L</em>

pOH = -log[OH⁻] = 1.22

pH = 14-pOH

pH = 12.78

4 0
3 years ago
Please help me urgently
e-lub [12.9K]
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3 years ago
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A sample of sugar contains 1.505 × 1023 molecules of sugar. How many moles of sugar are present in the sample?
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? mole ----------- 1.505 x10²³ molecules

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