We have a solution of NaOH and H₂CO₃
First, NaOH will dissociate into Na⁺ and OH⁻ ions
The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃
The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule
If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃
Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.
Explanation:
The reaction equation is as follows.

Initial : 0.3 0.4 0
Change: -x -x x
Equilbm: (0.3 - x) (0.4 - x) x
We know that, relation between standard free energy and equilibrium constant is as follows.

Putting the given values into the above formula as follows.


x = 0.1417
Hence, at equilibrium
= 0.1583 M
= 0.2583 M
Explanation:
To calculate the number of atoms in a sample, divide its weight in grams by the amu atomic mass from the periodic table, then multiply the result by Avogadro's number: 6.02 x 10^23.