What is the fraction of association (?) for the following potassium propionate solutions? Ignore activities. The Ka of propanoic acid is 1.34 × 10-5.
(a) 4.00 × 10-1 M K(C2H5CO2) ****the answer is not 4.31e-5
(b) 4.00 × 10-2 M K(C2H5CO2) ****the answer is not 1.36e-4
(c) 4.00 × 10-12 M K(C2H5CO2) ***** the answer is not 1.365e-1
The dissociation constant Ka represent the dissociation of acid as follows :
C2H5CO2H + H2O <-----------------> C2H5CO2^-1 + H3O^+1
This means for association of acid the value will be just inverse of this. So, for association/formation of acid the reaction can be written as:
K(C2H5CO2) + H2O <--------------> C2H5COOH + K^+1 + OH^-1
And for this the constant K can be written as:
K = 1/Ka = 1/(1.34×10^-5) = 7.463 × 10^4
And K = [C2H5CO2H][K^+1][OH^-1]/[K(C2H5CO2)]
Using ICE table
[K(C2H5CO2)] . [C2H5CO2^-1] . [OH-]. [K+]
I. 0.4. 0. 0. 0
C . -x . +x . +x . +x
E . 0.4 - x . x . x . x
Putting all the value in the expression
7.463×10^4 = x^3/(0.4 - x)
x = 0.399999
So, the concentration of associated K(C2H5CO2) = 0.4 - x
= 0.4 - 0.399999 = 0.000001
Fraction alpha = associated / total concentration
= 0.000001/0.4 = 2.5 ×10^-6
Similaly can be calculated for the other parts
What is the fraction of association (?) for the following potassium propionate solutions? Ignore activities. The Ka of...
What is the fraction of association (?) for the following potassium propionate solutions? Ignore activities. The Ka of propanoic acid is 1.34 × 10-5.(a) 5.00 × 10-1 M K(C2H5CO2)(b) 5.00 × 10-2 M K(C2H5CO2)(c) 5.00 × 10-12 M K(C2H5CO2)
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