Calculating pH of reaction mixture of silver nitrate and ascorbic acid

Question

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?

Answer 1

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 $\mathrm{pH}$.