Question

Review Col If K for NX3 is 1.5x10-6, what is the pOH of a 0.175 M aqueous solution of NX?! Express your answer numerically. View Available Hint(s) EVO AQ R O O ? pOH = Submit Part B If Kh for NX3 is 1.5x10-6, what is the percent ionization of a 0.325 M aqueous solution of NX;? Express your answer numerically to three significant figures. View Available Hint(s)

Answer 1

A)

NX3 dissociates as:

NX3 +H2O -----> NX3H+ + OH-

0.175 0 0

0.175-x x x

Kb = [NX3H+][OH-]/[NX3]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.5*10^-6)*0.175) = 5.123*10^-4

since c is much greater than x, our assumption is correct

so, x = 5.123*10^-4 M

So, [OH-] = x = 5.123*10^-4 M

use:

pOH = -log [OH-]

= -log (5.123*10^-4)

= 3.29

Answer: 3.29

B)

NX3 dissociates as:

NX3 +H2O -----> NX3H+ + OH-

0.325 0 0

0.325-x x x

Kb = [NX3H+][OH-]/[NX3]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.5*10^-6)*0.325) = 6.982*10^-4

since c is much greater than x, our assumption is correct

so, x = 6.982*10^-4 M

% dissociation = (x*100)/c

= 6.982*10^-4*100/0.325

= 0.2148 %

Answer: 0.215 %