Question

Determine the [H3O+] of a 0.170 M solution of formic acid.

Answer 1

Sol :-

ICE table is :

..........................HCOOH (aq)... .+... H₂O (l)---------------> HCOO^-(aq) ................+.............H₃O⁺(aq)

Initial (I).............0.170 M.................................................0.0 M..........................................0.0 M

Change (C)...........-y .......................................................+y.................................................+y

Equilibrium (E)....(0.170-y) M...........................................y M...............................................y M

Expression of K_a is :

K_a = [HCOO^-].[H₃O⁺] / [HCOOH]

1.8 x 10^-4 = y²/(0.170-y)

0.0000306 - 0.00018 y = y²

y² + 0.00018 y - 0.0000306 = 0

On solving, we have

y = 0.00544

So,

[H₃O⁺] = y = 5.44 x 10^-3 M