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.
Chemistry(Please help) - DrBob222, Saturday, April 14, 2012 at 11:09pm
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?
No. Kb = 1E-14/k2 for H2CO3.
How do I set up an ice table when I only have .15M?
To find the pH of the solution, we need to determine the concentration of the hydronium ion (H3O+).
Start by writing the reaction for the hydrolysis of carbonate ions:
CO3^2- + H2O ⇌ HCO3^- + OH^-
The equilibrium constant expression for this reaction is given by:
Kb = [HCO3^-][OH^-] / [CO3^2-]
We can assume that the concentration of CO3^2- is equal to the initial concentration of potassium carbonate, which is 0.15 M.
Let's denote x as the change in concentration of HCO3^- and OH^- ions from the initial concentration of CO3^2-. Therefore, at equilibrium, the concentration of HCO3^- and OH^- will both be equal to x.
Substituting these values into the equilibrium constant expression:
Kb = x * x / (0.15 - x)
Since the equilibrium constant value Kw = 1.0 x 10^-14 is given, and Kb = Kw / Ka (where Ka is the acid dissociation constant for carbonic acid H2CO3), we can rearrange the equation:
Kb = (1.0 x 10^-14) / Ka
Now we can solve for x. Rearrange the equation to isolate x:
x^2 = (1.0 x 10^-14) / Ka * (0.15 - x)
Since Ka for carbonic acid is approximately 4.45 x 10^-7, plug in these values:
x^2 = (1.0 x 10^-14) / (4.45 x 10^-7) * (0.15 - x)
Simplify the equation:
x^2 = (2.25 x 10^-8) * (0.15 - x)
Expand:
x^2 = 3.375 x 10^-9 - 2.25 x 10^-8 * x
Rearrange the equation:
x^2 + 2.25 x 10^-8 * x - 3.375 x 10^-9 = 0
Now you can solve this quadratic equation for x using the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
Plugging in the values:
x = [-(2.25 x 10^-8) ± sqrt((2.25 x 10^-8)^2 - 4(1)(-3.375 x 10^-9))] / (2)
Solve for x using a calculator, and you will obtain two values. Since we are interested in the concentration of H3O+, we choose the positive value for x.
Once you have determined the value of x, you can find the concentration of H3O+ (which is equal to the concentration of HCO3^- and OH^-) and use the pH equation:
pH = -log[H3O+]
Plug in the value of [H3O+] and calculate the pH.