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 K_a for these acids.

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

queen_honey_blossom answered 1 year ago

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