We’re asked to **determine the K**_{b}** **for a 0.140 M solution of a weak base with a pH of 11.25.

Recall that the Equilibrium reaction of any weak base, B in water is given by:

**B _{(aq)} + H_{2}O_{(l)} **

** (reactants) (products)**

• **B** →*weak base* → proton acceptor

• **H _{2}O** → will act as the

We write the **K _{b} expression for a weak base, B** as:

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\left[{\mathbf{HB}}^{\mathbf{+}}\right]\left[{\mathbf{OH}}^{\mathbf{-}}\right]}{\left[\mathbf{B}\right]}$

Since we’re dealing with a weak base and **K _{b} is an equilibrium expression **we will have to

A 0.140 M solution of a weak base has a pH of 11.25.

Determine K_{b} for the base.

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