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3 years ago

A 0.068 M solution of benzamide has a pOH of 2.91. What is the value of Kb for this compound?

Chemistry

1 answer:

igor_vitrenko [27]3 years ago

5 0

Answer:

Kb = 2.23 x 10⁻⁵

Explanation:

Benz-CO-NH₂ + H₂O => Benz-CO-NH₃⁺OH⁻ ⇄ Benz-CO-NH₃⁺ + OH⁻

given pOH = 2.91 => [OH⁻] = 10E-2.91 = 1.23 x 10⁻³M

Benz-CO-NH₃⁺OH⁻ ⇄ Benz-CO-NH₃⁺ + OH⁻

At Equilibrium 0.68M 1.23 x 10⁻³M 1.23 x 10⁻³M

Kb = [Benz-CO-NH₃⁺][OH⁻]/[ Benz-CO-NH₃⁺OH⁻]

= (1.23 x 10⁻³)² M²/ 0.068M = 2.23 x 10⁻⁵

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$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

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