The pH of a \(0.30 M\) solution of a weak base is 10.66 . What is the \(K_{\mathrm{b}}\) of the base?

Short Answer

Expert verified
The \(K_{\mathrm{b}}\) of the base is \(6.97 \times 10^{-8}\).

Step by step solution

01

Calculation of pOH

Since pH + pOH = 14 at 25°C, the pOH can be calculated as pOH = 14 - pH = 14 - 10.66 = 3.34.
02

Calculation of [OH^-] Concentration

Using the formula \([OH^-] = 10^{-\mathrm{pOH}}\), the concentration of OH^- can be obtained: \([OH^-] = 10^{-3.34} M = 4.57 \times 10^{-4} M\). This result implies that the concentration of the OH- is also the concentration of the weak base in its ionized form, \(B^-\).
03

Calculation of \(K_{\mathrm{b}}\)

We consider the auto-ionization of the weak base \(B\), \(B + H_2O \rightleftharpoons B^- + H_3O^+\). At equilibrium, the \([B^-]\) equals \([OH^-]\) and \([B]\) can be approximated to the initial concentration of the base, 0.30 M. Therefore, substituting these values into the \(K_{\mathrm{b}}\) expression \(K_{\mathrm{b}} = \frac{[B^-][OH^-]}{[B]}\) gives \(K_{\mathrm{b}} = \frac{(4.57 \times 10^{-4})^2}{0.30} = 6.97 \times 10^{-8}\).

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