Calculating pH of reaction mixture of silver nitrate and ascorbic acid

I am designing an experiment but before I do, I wanted to understand if the following method is right to calculate the expected $\mathrm{pH}$ when I mix $\pu{0.736 M}$ $\ce{AgNO3}$ and $\ce{C6H8O6}$.


I want the $\ce{AgNO3}$ to be the limiting reactant for this reaction. I think chemical equation will be: 
$$\ce{2AgNO3 + C6H8O6 <=> 2Ag (s) + C6H6O6 + 2HNO3},$$


which would mean that there would be 2 hydrogen ions.


$$\ce{2Ag+ + C6H8O6 <=> 2Ag (s) + C6H6O6 + 2H+}$$


If I start with $\pu{0.736M}$ $\ce{AgNO3}$ and I plan to have that in $\pu{1L}$ water solution (also intend the $\ce{C6H8O6}$ to be in $\pu{1L}$ water solution), then the moles of hydrogen generated will be:
$$\pu{0.736 mol} \ \ce{H+}$$ 


This would mean the concentration of $\ce{H+}$ will be 
$$\frac{\pu{0.736 mol}}{\pu{2L}} = \pu{0.368M}$$ 


So the $\mathrm{pH}$ of the solution would be $-\log 0.368 = 0.434$? This number does not seem right since ascorbic is a weak acid, is my logic correct? 




For solutions of weak acids and with simplifying assumptions [H+]=[A-], [A-] << [HA ], there is formula:


$$\mathrm{pH}=\frac12(\mathrm{p}K\_\mathrm{a} - \log {c})\tag{1}$$


For our case, if the stoichiometric ascorbic acid concentration $0.368/2=\pu{0.184 mol/L}$ is used:


$$\mathrm{pH}=\frac12(4.1 - \log {0.184}) \simeq 2.42\tag{2}$$


If reaction is completed, $\mathrm{pH}$ would be given by $\ce{HNO3}$ only.


**Situation would get complicated, if nitric acid starts to react with metallic silver or residual ascorbic acid. These side reactions would decrease mineral and eventually also organic acidity.**


$$\ce{3 Ag + 4 HNO3 -> 3 AgNO3 + 2 H2O + NO}$$
$$\ce{3 C6H8O6 + 2 HNO3 -> 3 C6H6O6 + 4 H2O + 2 NO}$$




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If more of the ascorbic acid is used, initial $\mathrm{pH}$ is again given by the above equation.


The ending $\mathrm{pH}$ is given by equation:


$$[\ce{H+}]= c\_1 + [\ce{A-}]\simeq K\_\mathrm{a}\frac{c\_2}{[\ce{A-}]}\tag{3}$$


where $c\_1$ is the nitric acid concentration, $c\_2$ is the residual ascorbic acid concentration. and the right equation side is taken from $(1)$


Concentration of $\ce{H+}$ is equal to sum of $\ce{HNO3}$ concentration $c\_1$ + $\ce{H+} from weak acid, that is equal to A- concentration ( HA -> H+ + A-). The latter is then approx equal to simplified extression derived from the acid dissociation constant.


$$c\_1 \cdot [\ce{A-}] + {[\ce{A-}]}^2 - K\_\mathrm{a}\cdot c\_2 \simeq 0 \tag{4} $$


$$[\ce{A-}]=\frac{-c\_1 + \sqrt{{c\_1}^2+4\cdot c\_1\cdot c\_2 \cdot K\_\mathrm{a}}}{2} \tag{5} $$


From (3):


$$[\ce{H+}]=\frac{c\_1 + \sqrt{{c\_1}^2+4\cdot c\_1\cdot c\_2 \cdot K\_\mathrm{a}}}{2} \tag{6} $$


$$[\ce{H+}]=c\_1 \cdot \frac{1 + \sqrt{1+ \frac{4\cdot c\_2 \cdot K\_\mathrm{a}}{c\_1}}}{2} \tag{7} $$


In the case by we use acid concentration twice as much as stoichiometric one, then $c\_1 = 2c\_2$. As $K\_\mathrm{a}=\pu{7.94e-5}$:


$$[\ce{H+}]=c\_1 \cdot \frac{1 + \sqrt{1+ 2 \cdot K\_\mathrm{a}}}{2} \simeq c\_1 \cdot 1.00004 \tag{8} $$


You can see the dissociation of the weak acid is almost suppressed by $\ce{H+}$ from nitric acid, so we can neglect the weak acid contribution to $\mathrm{pH}$.


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