..............HA ==>H^+ + A^-
initial....0.25M....0......0
change.......-x......x......x
equil......0.25-x....x.....x
If it is 3% dissociated, that means that x = 0.25*0.03 = ?
Substitute into the Ka expression and solve for Ka.
pH = -log(H^+)
In a .25 M solution, a weak acid is 3.0% dissociated.
A. calculate the pH of the solution
B. calculate the Ka of the acid
Please explain steps.
2 answers
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