We are being asked to calculate the percent ionization of a 0.325 M aqueous solution of NX_{3}.

Recall that the ** percent ionization **is given by:

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{ionization}}{\mathbf{=}}\frac{\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{[}\mathbf{B}{\mathbf{]}}_{\mathbf{initial}}}{\mathbf{\times}}{\mathbf{100}}}$

We know the initial concentration of NX_{3}, **0.325 M**. We just need to find the concentration of OH^{-} at equilibrium.

**NX _{3}** has a low K

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

• **NX**_{3} → *weak base* → proton acceptor

• **H _{2}O** → will act as the

**Equilibrium reaction:** **NX _{3(aq)} + H_{2}O_{(l)} **

If K_{b} for NX_{3} is 4.5×10^{−6}, what is the percent ionization of a 0.325 M aqueous solution of NX_{3}?

Express your answer numerically to three significant figures.

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