A weak base B (neutral charge) is dissolved in water at a concentration of 0.500 M....

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A weak base B (neutral charge) is dissolved in water at aconcentration of 0.500 M. The concentration of the resulting ionsis 0.170. What is the numerical value of Kb? The answer should bewritten to 4 decimal places and correct within +/-0.0020. Don'tforget that [B] is going to be less than 0.500 M.

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Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â BOH(aq) + H2O(l) --------------> B+(aq) + OH-(aq)

Â Â Â Â Â Â Â Â Â initialÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 0.5 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0 MÂ Â Â Â Â Â 0 M

Â Â Â Â Â Â Â Â Â changeÂ Â Â Â Â Â Â Â Â Â 0.17 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.17 MÂ Â Â Â 0.17M

Â Â Â equilibriumÂ Â Â Â Â Â Â Â Â Â 0.5-0.17Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.17Â Â Â Â Â Â Â Â Â Â Â Â 0.17

Â Â Â Â Â Â Â Â Â Kb= [B+][OH-]/[BOH]

Â Â Â Â Â Â Â Â Â Â Kb = 0.17^2/(0.5-0.17) = 0.0876

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