From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for...

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Chemistry

From the equilibrium concentrations given, calculate Ka for eachof the weak acids and Kb for each of the weak bases.
c) HCO2H
[HCO2H]= 0.524 M
[H3O+]= 9.8 x 10^-3 M
[HCO2-] 9.8 x 10^-3
d) C6H5NH3+
[C6H5NH3+]= 0.233
[C6H5NH2]= 2.3 x 10^-3
[H3O+]= 2.3 x 10^-3

Answer & Explanation Solved by verified expert

4.2 Ratings (760 Votes)

c) HCO2H

[HCO2H]= 0.524 M

[H3O+]= 9.8 x 10^-3 M

[HCO2-] 9.8 x 10^-3

Solution :-

HCO2H + H2O ----- > H3O+Â Â + HCOO-

0.524Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â +x

0.524-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 9.8 x 10^-3Â Â Â Â Â Â Â 9.8 x 10^-3

Ka= [H3O+][HCOO-]/[HCOOH]

Â Â Â Â = [9.8*10^-3][9.8*10^-3]/[0.524-9.8*10^-3]

Â Â Â = 1.87*10^-4

d) C6H5NH3+

[C6H5NH3+]= 0.233

[C6H5NH2]= 2.3 x 10^-3

[H3O+]= 2.3 x 10^-3

C6H5NH3+ + H2O ------- > C6H5NH2 + H3O+

0.233Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â +x

0.233-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2.3*10^-3Â Â Â 2.3*10^-3

Ka = [C6H5NH2][H3O+] /[C6H5NH3+]

Â Â Â Â = [2.3*10^-3][2.3*10^-3]/[0.233-2.03*10^-3]

Â Â Â Â = 2.29*10^-5

Kb= kw/ka

Â Â Â Â = 1*10^-14/2.29*10^-5

Â Â Â Â = 4.37*10^-10

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