A 625mL sample of an aqueous solution containing 0.275 mol propionic acid (CH3CH2CO2H),has [H30+]= 0.00239M.
What is the value of Ka for propionic acid?
Let HPr = propionic acid.
HPr ==> H^ + Pr^-
Set up an ICE chart.
Ka = (H^+)(Pr^-)/(HPr)
You know (H^+) from the problem. (Pr^-) is the same as (H^+). (HPr) = moles/L and you have the information to calculate that. Solve for Ka.
please help me to solve , I'm confused
I have get it now and the answer is
Ka =2.1×10^-5
To calculate the value of Ka for propionic acid (CH3CH2CO2H), we need to use the following equation:
Ka = [H3O+][C2H5COO-] / [C2H5CO2H]
From the information given, we are provided with the concentration of [H3O+] and the volume of the sample solution.
First, let's find the number of moles of propionic acid (C2H5CO2H) in the sample solution.
Number of moles = concentration x volume
Number of moles = 0.275 mol x (625 mL / 1000 mL)
Number of moles = 0.171875 mol
Since propionic acid is a weak acid, we can assume that the concentration of [C2H5COO-] (the acetate ion) is the same as the concentration of the conjugate base. Therefore, [C2H5COO-] = 0.171875 mol.
Now we have all the necessary information to calculate Ka.
Ka = [H3O+][C2H5COO-] / [C2H5CO2H]
Ka = (0.00239 M)(0.171875 mol) / 0.171875 mol
Ka = 0.00239 M
Therefore, the value of Ka for propionic acid is 0.00239 M.