Homework Answers
NH4Cl(aq) ------------> NH4^+ (aq) + Cl^- (aq)
0.18M ---------------- 0.18M
Kb of NH3 = 1.8*10^-5
----- NH3(aq) + H2O (l) --------------> NH4^+ (aq) + OH^- (aq)
I 0.056--------------------------------- 0.18 ------- 0
C -x ------------------------------------- +x -------------- +x
E 0.056-x --------------------------------- 0.18+x ------------- +x
Kb = [NH4^+][OH^-]/[NH3]
1.8*10^-5 = (0.18+x)(x)/(0.056-x)
1.8*10^-5*(0.056-x) = (0.18+x)(x)
x = 5.6*10^-6
[OH^-] = x = 5.6*10^-6M
POH = -log(5.6*10^-6)
= 5.25
PH = 14-POH
= 14-5.25
= 8.75 >>>>answer
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