Question

Problem 16.63

Calculate the percent ionization of hydrazoic acid (HN3) in solutions of each of the following concentrations (Ka is given in Appendix D in the textbook).

Ka=1.5*10^-9

Part B

0.113 M.

%=

Part C

4.30*10^-2

^%=

Answer 1

I will use HA to represent HN3
(see ka for HN3 is not in order of 10^-9 its in order of 10^-5 ,,Ka=1.3X10^-5 or 1.9X10^-5,still solving as given
PART B
HA <--> H^+1 + A^-
Ka = [H+][A-] / [HA] = 1.5 x 10^-9
X^2/(0.113-X) = 1.5x10^-9
NOTE
The general rule for whether or not you can drop the -x term is that if [HA]initial / Ka is greater than 100, you can drop the -x. In our case, [HN3]initial / Ka =0.113/1.5x10^-9=75333333.33>>100
so X^2 = 1.5x10^-9 X0.113
X=1.3019X10^-5 and 1.3019X10^-5/(0.113-1.3019X10^-5) = 0.011522% dissociated
similarly

PART C
X^2/(4.30*10^-2-X) = 1.5x10^-9
X =8.0311X10^-6 and 8.0311X10^-6/(4.3X10^-2-8.0311X10^-6) = 0.018680%
dissociated

X^2/(0.040-X) = 1.9x10^-5
if X << 0.040, then X = 0.00087 and 0.00087/(0.040-.00087) = 2.2%
dissociated

Without the assumption that X << original HA concentration, you would need to solve the quadratic equations. The answer so not vary significantly from those given above.