Asked by Cal21
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?
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?
Answers
Answered by
DrBob222
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.
.........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.