Calculate the hydronium ion concentration and pH in a 0.92 M solution of ammonium chloride.
Ka NH4+ = 5.6x10^-10
NH4+ dissociates as:
NH4+ -----> H+ + NH3
0.92 0 0
0.92-x x x
Ka = [H+][NH3]/[NH4+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.6*10^-10)*0.92) = 2.27*10^-5
since c is much greater than x, our assumption is correct
so, x = 2.27*10^-5 M
So, [H+] = x = 2.27*10^-5 M
use:
pH = -log [H+]
= -log (2.27*10^-5)
= 4.644
Answer:
[H+] = 2.27*10^-5 M
pH = 4.64
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