Question

Hey guys I need your help to solve the following question:
(a) 10^-25mol/L NaHCO3 and 10^-3 mol/L strong base NaOH are added to water (pKc1=6.3 and pkC2=10.4). What is the H2CO3 alkalinity of this solution.
(b) How could a solution with the same inorganic carbon species concentration (CT) and pH be made based on Na2CO3 reference species? Be exact and quantitative.
(c) What is the HCO3- alk of (b) solution.
(d) In what third way can (a) solution with the same CT and pH be made up? Again, be exact and quantitative.
I'm attaching my working for part (a) could you check it and help me. Thanks

Answer 1

Here's my working..

Answer 2

I have no idea what "H2CO3 alkalinity" of a solution means. Could you give me a definition?

Anyway, such a solution will yield:
2.16 mmol HCO3- and 1 mmol CO32-
pH=10.0

Answer 3

Hi @Vincent-Lyon.Fr,

Here's what I have as definition in my notes:
It's the amount of alkalinity added to reach the H2CO3 equivalence point of the solution.

Can you please explain to me how to do you come to the conclusion that the pH should be equal to 10?

Answer 4

From what I understand of your problem, you have one litre of sultion in which initial HCO3- is \(10^{-2.5}\) M and initial OH- is \(10^{-3}\) M.

Amounts in millimoles are:

\(HCO_3^-\) : 3.16 mmol
\(OH^-\) : 1.0 mmol

They will react according to:
\(HCO_3^-+OH^-\rightarrow CO_3^{2-}+H_2O\)

At equilibrium new amounts are:
\(HCO_3^-\) : 2.16 mmol
\(CO_3^{2-}\) : 1.0 mmol

all in same volume.

So \(pH=pK_{C2}+lg\Large \frac{[CO_3^{2-}]}{[HCO_3^-]}\normalsize =10.0\)