Question

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 is 1.9x10^-5

A).    0.383M

B).   0.113M

C).    3.66x10^-2

Answer 1

%ion = [H+]/M*100%

then

Ka = [H+][N3-]/[HN3]

solve for H+

[H+]^2 = Ka*[HN3]

[H+] = sqrt(Ka*[HN3])

then

%ion = [H+]/M*100%

%ion = (sqrt(Ka*[HN3]) )/M*100%

%ion = (sqrt((1.9*10^-5)*[HN3]) )/M*100%

a)

%ion = (sqrt(1.9*10^-5*[HN3]) )/M*100%

M = 0.383

%ion = sqrt((1.9*10^-5)*(0.383))/0.383*100% = 0.70433%

b)

%ion = (sqrt(1.9*10^-5*[HN3]) )/M*100%

M = 0.113

%ion = sqrt((1.9*10^-5)*(0.113))/0.113*100% = 1.29669%

c)

%ion = (sqrt(1.9*10^-5*[HN3]) )/M*100%

M = 3.66*10^-2

%ion = sqrt((1.9*10^-5)*(3.66*10^-2))/(3.66*10^-2)*100% = 2.2784%