Change in concentration when acids are diluted I came across this formula to calculate the resultant [H]+ concentration when two weak acids are mixed (Both acids of equal volumes are mixed together). 

The only problem i have with this is that, we are completely ignoring the fact that the concentration of both the acids are going to be halved after they are mixed together since, the volume is doubling. Is there anything i'm missing? The only way, this could be correct is if the concentrations that are given to me in the question are those after the acids have been mixed.

EDIT: This is the question i was talking about 


This is the solution i was given
Let suppose $c, c_1, c_2$ are concentrations of weak acids, and $K_\mathrm{a}, K_\mathrm{a1}, K_\mathrm{a2}$ are their respective acidity(dissociation) constants.

The equations $\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$ for a weak acid, respectively for 2 acids $\ce{[H+]}=\sqrt{K_\mathrm{a1} \cdot c_1 + K_\mathrm{a2} \cdot c_2}$

are approximately valid, if $c, (c_1, c_2) \gg K_\mathrm{a}(, K_\mathrm{a1}, K_\mathrm{a2}) \gg 10^-7 = \sqrt{K_\mathrm{w} }$, what also implies $\ce{[H+]} \gg 10^{-7}$.

Then we can approximate all $\ce{H+}$ comes from the acid dissociation, but OTOH, all acid is in non-dissociated form, leading to $$K_\mathrm{a} = \frac{\ce{[H+][A-]} }{\ce{[HA]}} \simeq \frac{{[H+]}^2 }{ c}$$ and $$\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$$

If, OTOH, $c \ll 10^{-7}=\sqrt{K_\mathrm{w}}$  and $c \ll K_\mathrm{a}$ we can approximate all acid is dissociated.

Then $$K_\mathrm{w} =  \ce{[H+][OH-]}$$

$$\ce{[H+]} - \ce{[OH-]} = c_1 + c_2$$

$$\ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}} = c_1 + c_2$$

$${\ce{[H+]}}^2 - (c_1 + c_2) \cdot \ce{[H+]} - K_\mathrm{w} = 0 $$

$$\ce{[H+]} = \frac{c_1 + c_2 \pm \sqrt {{(c_1 + c_2)}^2 + 4 \cdot K_\mathrm{w} }}{2}  $$

$$\ce{[H+]} = \frac{\pu{2.1e-9} \pm \sqrt {{(\pu{2.1e-9})}^2 + \pu{4e-14} }}{2} = \pu{1.0105e-7 mol/L }$$

$$\mathrm{pH} = 6.995$$

It is obvious, that if with full dissociation $\mathrm{pH}$ does not reach even the value $6.99$, the result of partial dissociation cannot be $6.7$