8.14 mmol/0.100L = 81.4 mM = 0.0814M
.........Na2SO3 + 2H2O ==> H2CO3 + 2NaOH
initial 0.0814M...............0.......0
change...-x..................x........x
equil...0.814-x..............x........x
Substitute into the Kc expression and solve for x.
Sodium carbonate (kc = 3.67 *10^-3) is a weak base that can react with two moles of water according to the following equilibrium:
Na2Co3 (aq) + 2 H2O yields H2CO3 + 2 NaOH
if 8.14 mMol of sodium carbonate is dissolved in 100ml of water and allowed to reach equilibrium, what will the concentration of its conjugate acid (H2CO3) be?
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