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?
Chemistry(Please help) - DrBob222, Tuesday, April 17, 2012 at 3:27pm
No. Kb = 1E-14/k2 for H2CO3.
How do I set up an ice table when I only have .15M?
.15M initial concentration of what? Or is it final concentration?
.........CO3^2- + HOH ==> HCO3^- + OH^-
initial..0.15...............0......0
change....-x................x......x
equil....0.15-x.............x......x
Kb = (Kw/K2) = (HCO3^-)(OH^-)/(CO3^2-)
To calculate the pH of an aqueous solution of potassium carbonate, you can follow these steps:
Step 1: Identify the relevant chemical reaction
In this case, we need to consider the hydrolysis reaction of carbonate ion (CO3^2-):
CO3^2- + H2O ⇌ HCO3^- + OH^-
Step 2: Set up an ICE table
An ICE table is used to keep track of the initial, change, and equilibrium concentrations of the species involved in the reaction. However, in this case, we don't have initial concentrations or changes because we are given the initial concentration of the compound, which is 0.15 M. Therefore, we only need to consider the equilibrium concentrations.
Species | CO3^2- | HCO3^- | OH^-
Initial | 0.15 M | 0 M | 0 M
Change | -x | +x | +x
Equilibrium | 0.15-x | x | x
Step 3: Write the expression for the base hydrolysis equilibrium constant (Kb)
The Kb expression for the reaction is Kb = ([HCO3^-] * [OH^-]) / [CO3^2-]
Step 4: Use the Kb expression to find x, the concentration of OH^-
Since we are given that the starting concentration of carbonate ion (CO3^2-) is 0.15 M and the reaction is 1:1, we can substitute the values into the Kb expression:
Kb = x^2 / (0.15 - x)
Step 5: Solve the quadratic equation to find x
Simplify the equation and solve for x using the quadratic equation formula or by factoring.
Step 6: Calculate the pOH and pH
Since pOH = -log[OH^-], we can calculate the pOH using the value of x.
Then, calculate the pH using the equation pH = 14 - pOH.
Please note that the concentrations obtained for HCO3^- and OH^- represent the initial concentration of each species at equilibrium.