To 1.0 L of a 0.35 M solution of HClO2 is added 0.18 mol of NaClO.
Calculate the [HClO2] at equilibrium.
How to solve this?
To solve this problem, we can set up an ICE (Initial, Change, Equilibrium) table for the reaction:
HClO2 + NaClO -> ClO2- + Na+ + H2O
Initial:
[HClO2] = 0.35 M
[NaClO] = 0.18 mol / 1.0 L = 0.18 M
[ClO2-] = 0 M (not part of the original solution)
[Na+] = 0 M (not part of the original solution)
[H2O] = 0 M (not part of the original solution)
Change:
Since the stoichiometry of the reaction is 1:1 for both HClO2 and NaClO, the change in concentration will be -0.18 M for both HClO2 and NaClO, and the concentrations of ClO2-, Na+, and H2O will increase by 0.18 M each.
Equilibrium:
[HClO2] = 0.35 - 0.18 = 0.17 M
Therefore, the equilibrium concentration of HClO2 is 0.17 M.