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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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A temperature of 50.9 °C is the same as ________ K.
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For each of the following unbalanced equations, calculate how many grams of each product would be produced by complete reaction
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The answer to your question is

Explanation:

Data

12.5 g of reactant

Balanced Reaction 1

                 TiBr₄ + 2H₂  ⇒  Ti  +  4HBr

Molar mass of TiBr₄ = 48 + (4 x 80) = 368 g

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Molar mass of HBr = 1 + 80 = 81

                 368 g of TiBr₄ ---------------- 48 g of Ti

                    12.5 g of TiBr₄ -------------- x

                    x = (12.5 x 48) / 368

                   x = 1.63 g of Ti

                  368 g of TiBr₄ ----------------4(81) g of HBr

                    12.5 g of TiBr₄ -------------  x

                    x = (12.5 x 324) / 368

                    x = 11 g of HBr

Balanced reaction 2

                  3SiH₄  +  4NH₃  ⇒   Si₃N₄  +  12H₂

Molar mass of SiH₄ = 28 + 4 = 32

Molar mass of Si₃N₄ = 28 x 3 + 14 x 4 = 84 + 56 = 140 g

Atomic mass of H₂ = 2 g

                  3(32) g of SiH₄ --------------- 140 g of Si₃N₄

                    12.5 g of SiH₄ --------------  x

                    x = 18.2 g of Si₃N₄

                  3(32) g of SiH₄ --------------- 24 g of H₂

                  12.5 g of SiH₄   --------------  x

                   x = 3.125 g of H₂

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4 years ago
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