If Kb for NX3 is 1.5x10-6 what is the percent ionization of a 0.325 M aqueous solution of NX3?

Question

Robert · Accepted Answer

NX3 +H2O ======== 
+NHX3 + 
-OH 
Kb  = [
+NHX3][
 -OH] / [NX3] 
Kb  = [x][x] / [NX3 – x] 
Kb  = 1.5 x 10
-6 
[NX3] = 0.325 M 
1.5 x 10-6  = [x][x] / [0.325 – x] 
If x in denominator is ignored compared to the concentration of NX3 
X2 = 1.5 x 10-6  x 0.325 = 4.875 x 10
-7 
X2 = 4.

If Kb for NX3 is 1.5x10-6 what is the percent ionization of a 0.325 M aqueous solution of NX3?

Solution

Answer To This Question Is Available To Download

Related Questions & Answers

Submit New Assignment