Find the pH of a 0.100 M solution of a weak monoprotic acid having Ka= 1.2×10^{−3}.
Express your answer to two decimal places.
Let the acid be HA
HA----> H+ +A-
Initial HA = 0.1 H+ = [A-]= 0
change [HA] =-x [H+] =[A-] =x
At Equilibrium HA= 1-x [H+] =[A-] =x
Ka= x^{2}/(1-x)= 1.2*10^{-3}
this equation gives when solved usong solved gives x =0.0341
[H+] =0.0341
pH= -log(0.0341)=1.467
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