An aqueous solution of carbonic acid reacts to reach equilibrium as described below.
H2CO3(aq) + H2O(l) <---> HCO3^-(aq) + H3O^+(aq)
The solution contains the following solute concentrations:
Carbonic Acid: 3.3*10^-2 mol/L
HCO3^-: 1.19*10^-4 mol/L
H3O^+: 1.19*10^-4 mol/L
Determine the Keq
Work:
Keq= (1.19*10^-4)(1.19*10^-4)/3.3*10^-2
Keq=1.4161*10^-8/3.3*10^-2
Keq= 4.29*10^-7
Needs a chemist
nevermind figured it out
To determine the equilibrium constant (Keq) for this reaction, you need to use the concentrations of the products and reactants in the equilibrium expression.
The equation for Keq is given by:
Keq = ([HCO3^-][H3O^+]) / [H2CO3][H2O]
Substituting the given concentrations:
Keq = (1.19 * 10^-4)(1.19 * 10^-4) / (3.3 * 10^-2)
Calculating the numerator:
(1.19 * 10^-4)(1.19 * 10^-4) = 1.4161 * 10^-8
Now calculating the denominator:
3.3 * 10^-2 = 3.3 * 10^-2
Dividing the numerator by the denominator:
Keq = 1.4161 * 10^-8 / (3.3 * 10^-2)
Keq = 4.29 * 10^-7
Therefore, the equilibrium constant (Keq) for this reaction is 4.29 * 10^-7.