Question

In: Chemistry

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

From the equilibrium concentrations given, calculate Ka for each of 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

Expert Solution

c) HCO2H

[HCO2H]= 0.524 M

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

[HCO2-] 9.8 x 10^-3

Solution :-

HCO2H + H2O ----- > H3O+ + HCOO-

0.524 0 0

-x +x +x

0.524-x 9.8 x 10^-3 9.8 x 10^-3

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

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

= 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+

0.233 0 0

-x +x +x

0.233-x 2.3*10^-3 2.3*10^-3

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

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

= 2.29*10^-5

Kb= kw/ka

= 1*10^-14/2.29*10^-5

= 4.37*10^-10

queen_honey_blossom answered 2 years ago

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