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What is the Kb value for a 0.20M solution with a 3.0% ionization? NH3 + H2O <—> NH4^+ + OH^-

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Answer to a math question What is the Kb value for a 0.20M solution with a 3.0% ionization? NH3 + H2O <—> NH4^+ + OH^-

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107 Answers

To find the

K_b

value for the given solution, we first need to calculate the concentration of

OH^-

produced using the initial concentration of

NH_3

and the ionization percentage.
Given:
- Initial concentration

[NH_3]

= 0.20 M
- Ionization percentage = 3.0%

Concentration of

OH^-

\text{Concentration of } OH^- = 0.20 \, \text{M} \times \frac{3.0}{100} = 0.006 \, \text{M}

Since each molecule of

NH_3

produces one

NH_4^+

ion and one

OH^-

ion, we have:

K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = \frac{0.006 \times 0.006}{0.20} = 1.8 \times 10^{-4}

Therefore, the

K_b

value for the given 0.20 M solution of

NH_3

with a 3.0% ionization is approximately

1.8 \times 10^{-4}

\boxed{K_b \approx 1.8 \times 10^{-4}}

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What is the Kb value for a 0.20M solution with a 3.0% ionization? NH3 + H2O <—> NH4^+ + OH^-

Answer to a math question What is the Kb value for a 0.20M solution with a 3.0% ionization? NH3 + H2O <—> NH4^+ + OH^-

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