Question

In: Chemistry

1. In the laboratory, a general chemistry student measured the pH of a 0.538 M aqueous...

1. In the laboratory, a general chemistry student measured the pH of a 0.538 M aqueous solution of nitrous acid to be 1.792.  
She then measured the pH of a 0.538 M aqueous solution of hydrofluoric acid to be 1.722.  
Use the information she obtained to determine the Ka for these acids.

Solutions

Expert Solution

PH   = 1.792

-log[H^+]   = 1.792

    [H^+]     = 10^-1.792   = 0.01614M

at equilibrium [H^+]   = [NO2^-]

           HNO2(aq) -------------> H^+ (aq) + NO2^- (aq)

I         0.538                            0                   0

C       -0.01614                      0.01614          0.01614

E         0.52186                     0.01614          0.01614

     Ka   =   [H^+][NO2^-]/[HNO2]

             = 0.01614*0.01614/0.52186   = 0.00049   = 4.9*10^-4 >>>>answer

part-B

PH   = 1.722

-log[H^+]   = 1.722

    [H^+]     = 10^-1.722   = 0.01896M

at equilibrium [H^+]   = [F^-]

           HF(aq) -------------> H^+ (aq) + F^- (aq)

I         0.538                            0                   0

C       -0.01896                      0.01896         0.01896

E         0.51904                    0.01896          0.01896

     Ka   =   [H^+][F^-]/[HF]

             = 0.01896*0.01896/0.51904    = 0.00069   = 6.9*10^-4 >>>>>answer


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