Calculate the Ka of a weak acid if a 0.14626M solution of it has a pH of 6.32.
Hint: Answer includes 3 significant figures.
let the weak acid be HX , it ionizes as
HX--> H+ + X-
Ka= [H+][X-]/[HX}
given
pH = 6.32 = -log[H+]
[H+] = 10^-pH = 4.786 x10^-7
[X-] =4.786 x10^-7
[HX] = 0.14626 M
Ka = 4.786 x10^-7 * 4.786 x10^-7/0.14626
= 156.6 x10^-14
Answer Ka = 1.57 x10^-12
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Calculate the Ka of a weak acid if a 0.14626M solution of it has a pH...
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