Calculate the pH of an aqueous solution of 0.15 M potassium carbonate.
I know that pH = -log(H30+) but I am not sure how to start this problem.
Hydrolyze the CO3^2-.
CO3^- + HOH ==> HCO3^- + OH^-
Do an ICE chart, Kb = Kw/k2 for H2CO3.
so for kb I do 1e-14/.15?
To calculate the pH of an aqueous solution of potassium carbonate, you need to consider the dissociation of the compound in water. Potassium carbonate (K2CO3) dissociates into potassium ions (K+) and carbonate ions (CO3^2-).
Step 1: Write the balanced chemical equation for the dissociation of potassium carbonate:
K2CO3(s) → 2K+(aq) + CO3^2-(aq)
Step 2: Identify the relevant equilibrium expression. Since potassium carbonate only partially dissociates in water, we can use an equilibrium constant called the dissociation constant (Kd) to express the extent of dissociation:
Kd = [K+][CO3^2-] / [K2CO3]
Step 3: Determine the concentration of carbonate ions (CO3^2-) in the solution. In this case, we have a 0.15 M potassium carbonate solution. Since potassium carbonate dissociates in a 1:1 ratio, the concentration of carbonate ions is also 0.15 M.
Step 4: Apply the equilibrium expression and calculate the concentration of potassium ions ([K+]):
Kd = [K+][CO3^2-] / [K2CO3]
Kd = [K+][0.15] / 0.15
1 = [K+]
Step 5: Calculate the concentration of hydrogen ions ([H3O+]). Since the dissociation of potassium ions does not contribute any hydrogen ions to the solution, the concentration of hydrogen ions will be negligible.
Step 6: Calculate the pH using the equation:
pH = -log[H3O+]
pH = -log(10^-14) (since [H3O+] is negligible)
pH = -(-14)
pH = 14
Therefore, the pH of the aqueous solution of 0.15 M potassium carbonate is 14.