$V_1$ litres of a strong acid $HB_1$ with initial concentration $c_1$ are mixed with $V_2$ litres of a strong acid $HB_2$ with initial concentration $c_2$ in order to obtain a mixture with volume $V=V_1+V_2$. Each mole of strong acid liberates one mole of $H_3O^+$: $n_1$ $ =$ $ n_{H_3O^+}$ liberated by $HB_1$ $=$ $c_1\cdot V_1$ $n_2 $ $=$ $ n_{H_3O^+}$ liberated by $HB_2$ $=$ $c_2\cdot V_2$ $pH $ $=$ $ -log\frac{n_{H_3O^+}}{V}$ thus:
Mixture of $V_1$ litres of a strong acid $HB_1$ with initial concentration $c_1$ with $V_2$ litres of a strong acid $HB_2$ with initial concentration $c_2$ : $pH$ $ =$ $-log\frac{n_1+n_2}{V_1+V_2}$ $pH $ $=$ $-log\frac{c_1V_1+c_2V_2}{V_1+V_2}$
$V_1$ litres of a strong acid $HB_1$ with initial concentration $c_1$ are mixed with $V_2$ litres of a weak acid $HB_2$ with initial concentration $c_2$ in order to obtain a mixture with volume $V=V_1+V_2$. Each mole of strong acid releases a mole $H_3O^+$, we neglect $H_3O^+$ released by the weak acid: $n_1 $ $=$ $ n_{H_3O^+}$ liberated by $HB_1$ $=$ $c_1\cdot V_1$ $pH $ $=$ $ -log\frac{n_{H_3O^+}}{V}$ thus:
Mixture of $V_1$ litres of a strong acid $HB_1$ with initial concentration $c_1$ with $V_2$ litres of a weak acid $HB_2$ with initial concentration $c_2$ in order to obtain a mixture with volume $V=V_1+V_2$.: $pH$ $ =$ $-log\frac{n_1}{V_1+V_2}$ $pH $ $=$ $-log\frac{c_1V_1}{V_1+V_2}$
$V_1$ litres of a weak acid $HB_1$ (acidity constant $K_{a1}$) with initial concentration $c_1$ are mixed with $V_2$ litres of a weak acid $HB_2$ with initial concentration $c_2$, (acidity constant $K_{a2}$) in order to obtain a mixture with volume $V=V_1+V_2$.. We need the concentrations of the two acids $c_{1m}$ and $c_{2m}$ 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 acid $HB_1$ (acidity constant $K_{a1}$) with initial concentration $c_1$ are mixed with $V_2$ litres of a weak acid $HB_2$ with initial concentration $c_2$ (acidity constant $K_{a2}$): $pH$ $ = $ $-\frac{1}{2}log(K_{a1}c_{1m}$ $+$ $K_{a2}c_{2m})$
Example 1: Mixture of $1.0 \;L \; HCl \;0.10\;M $ with $0.50 \;L \; HBr \;0.30\;M $: $pH$ $ =$ $-log\frac{0.1\cdot1.0+0.30\cdot0.50}{1.0+0.5}$ = $0.70 $
Example 2: Mixture of $20.0 \;mL \; HCl \;0.50\;M $ with $60.0 \;mL \; CH_3COOH \;0.05\;M $: $pH$ $=$ $-log\frac{0.50\cdot0.02}{0.02+0.06}$ = $0.90 $
Example 3: Mixture of $25.0 \;mL \; HCOOH \;0.10\;M $ with $50.0 \;mL \; CH_3COOH \;0.01\;M $: $pH$ $ =$ $-\frac{1}{2}log(K_{a1}c_{1m}+K_{a2}c_{2m})$ = $-\frac{1}{2}log(10^{-3.75}\frac{0.10\cdot 0.025}{0.025+0.050}$ $+$ $10^{-4.75}\frac{0.01\cdot 0.050}{0.025+0.050})$ = $2.60 $