1) HCOOH(aq) <-----------> HCOO-(aq) + H+(aq)
Initailly, [HCOOH] = x M & [H+] = [HCOO-] = 0 M
Let at eqb., [HCOOH] = (x-y) M & [H+] = [HCOO-] = y M
Now, pH = -log[H+] = -logy
Thus, 3.26 = -logy
or, y = 10-3.26 = 5.5*10-4 M
Hence , Ka = {[H+]*[HCOO-]}/[HCOOH]
or, y2/(x-y) = 1.7*10-4
or, x = 1.23*10-3 M
9) A formic acid, HCOOH (Ka = 1.7 x 10^-4), solution has a pH of 3.26....
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