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?
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.
To find the concentration of the conjugate acid (H2CO3), we need to calculate the equilibrium concentration using the given information.
First, let's calculate the moles of sodium carbonate (Na2CO3) in the solution:
Moles = concentration (in mol/L) * volume (in L)
Given concentration = 8.14 mMol = 8.14 × 10^-3 mol/L
Volume = 100 mL = 100 × 10^-3 L
Moles of Na2CO3 = (8.14 × 10^-3 mol/L) * (100 × 10^-3 L) = 0.814 mmol
According to the balanced equation, Na2CO3 reacts with two moles of water to form one mole of H2CO3. Therefore, the moles of H2CO3 formed will also be 0.814 mmol.
Now, we need to calculate the concentration of H2CO3.
Concentration = Moles / volume (in L)
Volume = 100 mL = 100 × 10^-3 L
Concentration of H2CO3 = (0.814 mmol) / (100 × 10^-3 L) = 8.14 mmol/L
So, the concentration of the conjugate acid (H2CO3) will be 8.14 mmol/L.