..........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.
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.
2 answers
0.0185*0.0155=2.8675*10-4
2.8675*10-4/0.0185M=1.55*10-3
2.8675*10-4/0.0185M=1.55*10-3
2 answers