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Simplified calculation of the pH of weak bases

Condition of simplification

We know that the formula: $K_b=\frac{[OH^-]^2}{c_B-[OH^-]}$ is used to calculate $[OH^-]$, then the pH of a dilute solution of a weak base. The use shows that often $[OH^-]$ is much smaller than $c_b$: Provided $[OH ^ -]$ is more than 100 times smaller than $c_b$ , we can neglect $[OH^-]$ in the expression $c_B-[OH ^ -]$ and replace this expression by $c_b$. Let us examine this condition: $[OH^-]$ is more than 100 times smaller than $c_B$ $[OH^-]$ $\lt$ $ \frac{c_B}{100}$ $\frac{(\frac{c_B}{100})^2}{c_B-\frac{c_B}{100}}$ $\gt$ $ K_a$ $\frac{c_B}{9900}$ $\gt$ $ K_b$ $\frac{K_b}{c_B}\lt 1,01\cdot 10^{-4}$ $\approx $ $10^{-4}$

Simplified formula

Simplifying under that condition, we can neglect $[OH^-]$ in the expression $c_B-[OH^-]$ and so we have: $K_b$ $=$ $\frac{[OH^-]^2}{c_B-[OH^-]} $ $\approx$ $ \frac{[OH^-]^2}{c_B}$ therefore:

$\frac{K_b}{c_B}\lt 10^{-4}$ $\Rightarrow$ $[OH^-]$ $=$ $\sqrt{K_bc_B}$ $pOH$ $=$ $\frac{1}{2}pK_b-\frac{1}{2}log(c_B)$

Example: Given a solution of a weak base $10^{-1}\frac{mol}{L}$ with $K_b=10^{-6}$: As $\frac{K_b}{c_B}=10^{-5}\lt 10^{-4}$, we have: $pH$ $=$ $14-\frac{1}{2}6+\frac{1}{2}log10^{-1}$ $=$ $10,5$