Asked by cindy

A 0.185 M solution of a monoprotic acid has a percent ionization of 1.55%. Determine the acid ionization constant (Ka) for the acid.

Answers

Answered by DrBob222
..........HA ==> H^+ + A^-
initial..0.185...0.....0
change....-x.....x.....x
equil...0.185-x..x.....x

x = 0185(0.0155) = ?
Then substitute into the Ka expression and solve for Ka.
Answered by Gerhard
0.0185*0.0155=2.8675*10-4
2.8675*10-4/0.0185M=1.55*10-3
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