$V_1$ litres of a strong base $B_1$ with initial concentration $c_1$ are mixed with $V_2$ litres of a strong base $B_2$ with initial concentration $c_2$ in order to obtain a mixture with volume $V=V_1+V_2$. Each mole of strong base liberates one mole of $OH^-$: $n_1$ $ =$ $ n_{OH^-}$ liberated by $B_1$ $=$ $c_1\cdot V_1$ $n_2$ $ = $ $n_{OH^-}$ liberated by $B_2$ $=$ $c_2\cdot V_2$ $pH$ $ =$ $ 14$ $+$ $log\frac{n_{OH^-}}{V}$ thus:
Mixture of $V_1$ litres of a strong base $B_1$ with initial concentration $c_1$ with $V_2$ litres of a strong base $B_2$ with initial concentration $c_2$: $pH$ $ =$ $14+log\frac{n_1+n_2}{V_1+V_2}$ $pH $ $=$ $14+log\frac{c_1V_1+c_2V_2}{V_1+V_2}$
$V_1$ litres of a strong base $B_1$ with initial concentration $c_1$ are mixed with $V_2$ litres of a weak base $B_2$ with initial concentration $c_2$ in order to obtain a mixture with volume $V=V_1+V_2$. Each mole of strong base liberates one mole of $OH^-$, $OH^-$ liberated by the weak base is neglected: $n_1$ $=$ $ n_{OH^-}$ liberated by $B_1$ $=$ $c_1\cdot V_1$ $pH$ $ =$ $ 14+log\frac{n_{OH^-}}{V}$ thus:
Mixture of $V_1$ litres of a strong base $B_1$ with initial concentration $c_1$ with $V_2$ litres of a weak base $B_2$ with initial concentration $c_2$: $pH$ $ =$ $14+log\frac{n_1}{V_1+V_2}$ $pH $ $=$ $14+log\frac{c_1V_1}{V_1+V_2}$
$V_1$ litres of a weak base $B_1$ with initial concentration $c_1$ (basicity constant $K_{b1}$) are mixed with $V_2$ litres of a weak base $B_2$ with initial concentration $c_2$ (basicity constant $K_{b2}$) in order to obtain a mixture with volume $V=V_1+V_2$. We need the concentrations $c_{1m}$ and $c_{2m}$ of the two bases in the mixture: $c_{1m}= \frac{n_1}{V}=\frac{c_1V_1}{V_1+V_2}$ $c_{2m}= \frac{n_2}{V}=\frac{c_2V_2}{V_1+V_2}$ Assume without proof:
Mixture of $V_1$ litres of a weak base $B_1$ with initial concentration $c_1$ (basicity constant $K_{b1}$) with $V_2$ litres of a weak base $B_2$ with initial concentration $c_2$ (basicity constant $K_{b2}$) $pH $ $ = $ $14+\frac{1}{2}log(K_{b1}c_{1m}$ $ + $ $K_{b2}c_{2m})$
Example 1: Mixture of $0.01 \;mole \; CH_3CH_2ONa $ (solid, volume can be neglected) with $0.02 \;L \; NaOH \;0.10\;M $: $pH$ $ =$ $14$ $+$ $log\frac{0.01+0.10\cdot0.02}{0.02}$ = $13.78 $
Example 2: Mixture of $20.0 \;mL \; NaOH \;0.25\;M $ with $60.0 \;mL \; NH_3 \;0.01\;M $: $pH$ $ =$ $14+log\frac{0.25\cdot0.02}{0.02+0.06}$ = $12.8 $
Example 3: Mixture of $25.0 \;mL \; NH_3 \;0.10\;M $ with $50.0 \;mL \; C_2H_5NH_2 \;0.05\;M $: $pH$ $ =$ $14$ $+$ $\frac{1}{2}log(K_{b1}c_{1m}+K_{b2}c_{2m})$ = $14$ $+$ $\frac{1}{2}log(10^{-4.80}\frac{0.10\cdot 0.025}{0.025+0.050}$ $+$ $10^{-3.33}\frac{0.05\cdot 0.050}{0.025+0.050})$ = $11.60 $