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A 0.513 M solution of a weak base has a pH of 11.4. What is the Kb of the base?

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2.64*10^-5

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We will assume that this is monobasic and the temp is 25C so that Kw=10^-14

If pH is 11.4 then pOH is 14-11.4 = 2.6

so [OH-] = 10^-2.6 = 2.512 x 10^-3

assuming that [weak base] at equilibrium is large compared with [OH-] then

Kb is approximately [OH-]^2/[weak base]

Kb=(2.512 x 10^-3)^2/0.513

=1.23 x 10^-5

but check my maths.

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