Given:
M(HCl) = 0.45 M
V(HCl) = 10 mL
M(NH3) = 0.45 M
V(NH3) = 10 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.45 M * 10 mL = 4.5 mmol
mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 0.45 M * 10 mL = 4.5 mmol
We have:
mol(HCl) = 4.5 mmol
mol(NH3) = 4.5 mmol
4.5 mmol of both will react to form NH4+ and H2O
NH4+ here is strong acid
NH4+ formed = 4.5 mmol
Volume of Solution = 10 + 10 = 20 mL
Ka of NH4+ 5.6*10^-10
concentration ofNH4+,c = 4.5 mmol/20 mL = 0.225 M
NH4+ + H2O -----> NH3 + H+
0.225 0 0
0.225-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)*0.225) = 1.123*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.123*10^-5 M
[H+] = x = 1.123*10^-5 M
use:
pH = -log [H+]
= -log (1.123*10^-5)
= 4.9498
Answer:
[H3O+] = 1.12*10^-5 M
pH = 4.950
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