NH4Cl(aq) -------------> NH4^+ (aq) + Cl^- (aq)
0.0321M 0.0321M
NH4^+ (aq) + H2O(l) --------------> NH3(aq) + H3O^+ (aq)
I 0.0321 0 0
C -x +x +x
E 0.0321-x +x +x
Ka = Kw/Kb
= 1*10^-14/(1.75*10^-5)
= 5.7*10^-10
Ka = [NH3][H3O^+]/[NH4^+]
5.7*10^-10 = x*x/(0.0321-x)
5.7*10^-10*(0.0321-x) = x^2
x = 4.27*10^-6
[H3O^+] = x = 4.27*10^-6M
PH = -log[H3O^+]
= -log4.27*10^-6
= 5.3696>>>>>answer
the Kb of ammonia (NH3) at 25.0 C is 1.75 x 10-5. Calculate the pH of...
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