The water in a carbonated drink is the A solute B soultion C solvent D mixture
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Answer: Concentration of = 0.328 M
Concentration of = 0.328 M
Concentration of = 0.532 M
Explanation:
Moles of and
= 0.430 mole
Volume of solution = 0.500 L
Initial concentration of and
=
The given balanced equilibrium reaction is,
Initial conc. 0.860M 0.860M 0
At eqm. conc. (0.860-x) M (0.860-x) M (x) M
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 0.532 M
Thus, the concentrations of at equilibrium are :
Concentration of = (0.860-x) M =(0.860-0.532) M = 0.328 M
Concentration of = (0.860-x) M = (0.860-0.532) M = 0.328 M
Concentration of = x M = 0.532 M