I have a collection of old books (all 80+ years old), recently, I received a British Chemistry text from 1903 (intro page below) - this being the oldest book of my collection (112 years old at the time of writing this question):
One thing I notice is that there is a distinct smell that comes ...
Basically I'm titrating $H^+$ (concentration .01 M, volume 2 mL) by some $CO_3^{2-}$ and I wanna find the volumes at which pH=pKa for $H_2CO_3$'s two pKas
Basically there's a first reaction $CO_3^{2-}+2H_3O^+=H_2CO_3+2H_2O$ with a constant $\frac{1}{K_{a1}K_{a2}}$ followed by $CO_3^{2-}+H_2CO_3=2HCO_3^-$ with a constant $\frac{K{a_1}}{K_{a_2}}$
(btw, do you know any software to simulate such titrations ? I have one in French that's very good but for some weird reason it doesn't know carbonic acid)
> The long-held belief that carbonic acid could not exist as a pure compound has reportedly been recently disproved by the preparation of the pure substance in both solid and gas form by University of Innsbruck researchers.
Basically I'm titrating $H^+$ (concentration .01 M, volume 2 mL) by some $CO_3^{2-}$ and I wanna find the volumes at which pH=pKa for $H_2CO_3$'s two pKas
then pH=pK_a1 (the lower one) should be at half the volume of the first inflection point and pH=pk_a2 (the higher one) should be at 3/4 of the second inflection point resp. in the middle between the first and the second inflection point
The first reaction is not simply of the kind $A^{2-}+H^+=AH$, it's a bit more complex, hence that formula doesn't apply here
The first reaction has a constant $\frac{1}{K_1K_2}$ and the second has a constant $\frac{K_1}{k_2}$ where $K_1$ is the constant for $H_2A=H^++HA^-$ and $K_2$ for $HA^-=H^++A^{2-}$
Because as I said above, your reactions occur at the same time (since the pKas have a difference under 4), whereas mine occur one after another (since their pKa have a difference greater than 4)
"Because as I said above, your reactions occur at the same time" ... then we will not come to a solution as I think that it is exactly the other way around
but I'm far from omniscient and so you might be right
What makes you think that the two reactions you initially wrote are successive ? The rule I know is that one can say that they are successive when their difference in pKa is greater than 4
my reactions cannot occur (more or less) simultaneously (then your reactions) as the hydrogencarbonate that the second reaction needs, need to be produced by the first reaction. the same is for your reactions, where you need the carbonic acid from the second step to be produced by the first step
there are exactly two steps you have to consider $$\ce{H+ + CO_3^2- -> HCO_3^-}$$ and $$\ce{H+ + HCO_3^- -> H2CO3 -> H2O + CO2}$$
For that second reaction to occur, you don't need any additional $CO_3^{2-}$ from the burrette to be introduced; it occurs naturally right after the first one
Hence those two reactions will only correspond to one inflection in the curve
in this case you have to "look" at it from the end ... think of it as being the carbonate that is titrated with hydrochloric acid ... then the graph would be mirrored at 1.5 mL about the Y axis
One thing that that has already been mentioned in the comments is the impact of $\ce{CO2}$ on the equilibrium. If you look at the following figure you can see the pH diagram of carbonic acid.
The equilibrium between $\ce{CO2}$ and $\ce{H2CO3}$ $$\ce{CO2 + H2O <=> H2CO3}$$ depends very much on ...
I have a collection of old books (all 80+ years old), recently, I received a British Chemistry text from 1903 (intro page below) - this being the oldest book of my collection (112 years old at the time of writing this question):
One thing I notice is that there is a distinct smell that comes ...
One thing that that has already been mentioned in the comments is the impact of $\ce{CO2}$ on the equilibrium. If you look at the following figure you can see the pH diagram of carbonic acid.
The equilibrium between $\ce{CO2}$ and $\ce{H2CO3}$ $$\ce{CO2 + H2O <=> H2CO3}$$ depends very much on ...
The Periodic Table
The Periodic Table