In: Chemistry
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.
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