1st calculate the volume of HCl required
At equivalence point,
mol of methylamine = mol of acid
0.195 M * 30.0 mL = 0.305 M * V
V = 19.2 mL
This is volume of acid added
CH3NH2 + HCl <—> CH3NH3+
mol of CH3NH3+ formed = mol of CH3NH2
= 0.195 M * 30.0 mL
=5.85 mmol
[CH3NH3+] = number of mol / total volume
= 5.85 mmol / (30.0 + 19.2) mL
= 0.119 M
Now this is weak acid
Ka = 10^-14/Kb
= 10^-14 / (4.4*10^-4)
= 2.27*10^-11
It will dissociate as,
CH3NH3+ <—> CH3NH2 + H+
0.119 0 0 (initial)
0.119-x x x (at equilibrium)
Ka = x*x/(0.119-x)
2.27*10^-11 = x*x/(0.119-x)
since Ka is small, x will be small and it be ignored as compared to 0.119
above expression thus becomes,
2.27*10^-11 = x*x/(0.119)
x = 1.64*10^-6 M
So,
[H+] = x = 1.64*10^-6 M
pH = -log [H+]
= -log (1.64*10^-6)
= 5.78
Answer: 5.78
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