Write the equation for the dissociation of carbonic acid.
H2CO3 + H2O <---> HCO3− + H3O+
carbonic acid is going to be 1M (mol dm-3) 20 ml
How much water in ml should I have to full dissociate 20 ml of 1M carbonic acid? To determine pKa... i was going to find the pH of the solution after it has been dissciated (carbnoic acid is a weak acid, will not fully dissociate).
from the pH... i will find the [H+] and then plug in the formula
Ka = [HCO3−] x [H3O+] / [H2CO3]
This is my last resort i don't have any thing else but done this help at all?
The dissociation of H2CO3 occurs in two steps.
H2CO3 + H2O ==> H3O^+ + HCO3^-
HCO3^- + H2O ==> H3O^+ + CO3^2-
not really I am just trying to understand what the chemical equation looks like
Well, well, well, looks like we're diving into the world of chemistry, huh? Alright, hold on to your test tubes, I'm about to drop some knowledge on you.
The equation for the dissociation of carbonic acid is:
H2CO3 ⇌ H+ + HCO3-
In other words, when carbonic acid breaks up, it forms a hydrogen ion (H+) and a bicarbonate ion (HCO3-). It's like carbonic acid is saying, "See ya later, guys! I'm splitting up and going my own way!" Ah, chemistry, always keeping things interesting.
To write the equation for the dissociation of carbonic acid (H2CO3), we need to consider the ionization or dissociation reaction it undergoes when dissolved in water. Carbonic acid is a weak acid, meaning it partially dissociates in water.
The equation for the dissociation of carbonic acid can be written as follows:
H2CO3 ⇌ H+ + HCO3-
In this equation, H2CO3 represents the carbonic acid molecule, H+ is the hydrogen ion (proton), and HCO3- is the bicarbonate ion, which is the conjugate base formed after the loss of a hydrogen ion.
It's important to note that the double-headed arrow (⇌) indicates that the reaction can proceed in both directions, indicating the equilibrium between the reactants and the products.
Well:
pKa = -log Ka, and
Ka = [H+][A-] / [HA]
Therefore Ka = [H+][HCO3-] / [H2CO3]
[H+] = Ka . [H2CO3] / [HCO3-]
does this help?