# 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?

• If you start equimolar and the reaction goes to completion, there is no more ascorbic acid. What pH do you expect at the beginning of the reaction?
– Karsten
Apr 25, 2020 at 3:48
• To rephrase Karsten's comment - The acid build up in the solution is the result of a redox reaction where ascorbic acid is oxidized to dehydroascorbic acid, not the dissociation of ascorbic acid.
– MaxW
Apr 25, 2020 at 4:10
• That is quite a lot of silver. Is that experiment worthy of it ? Apr 25, 2020 at 4:20
• Would the pH of ascorbic acid when added to water be the pH in the reaction with silver nitrate? Is that what you mean @MaxW Apr 25, 2020 at 4:49
• user510 - If you don't add any other acid, then yes the initial pH would be due to the dissociation of ascorbic acid.
– MaxW
Apr 25, 2020 at 4:52

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.

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

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 $$\mathrm{pH}$$. • Could you explain how eq (3) was derived Apr 26, 2020 at 16:40 • Ok so 0.1 M$\ce{AgNO3}$will have pH then of essentially -log(0.1) = 1? Apr 26, 2020 at 18:47 • But if it reacts with ascorbic acid the pH will still be close to 1 which you described above because the H+ concentration will be dominated by nitric acid Apr 26, 2020 at 18:50 • Ok I see. I am going to add 1.25 times stoichiometry concentration (to make sure silver nitrate is limiting) so I do not think that will drive up the pH of the initially$\ce{AgNO3}$solution Apr 26, 2020 at 18:58 • is c1 =$\frac{\sqrt {0.736 x K_b}}{Volume}\$? Apr 28, 2020 at 23:56