find the volume of HCl used to reach equivalence point
M(CH3NH2)*V(CH3NH2) =M(HCl)*V(HCl)
0.175 M *40.0 mL = 0.315M *V(HCl)
V(HCl) = 22.2222 mL
Given:
M(HCl) = 0.315 M
V(HCl) = 22.2222 mL
M(CH3NH2) = 0.175 M
V(CH3NH2) = 40 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.315 M * 22.2222 mL = 7 mmol
mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)
mol(CH3NH2) = 0.175 M * 40 mL = 7 mmol
We have:
mol(HCl) = 7 mmol
mol(CH3NH2) = 7 mmol
7 mmol of both will react to form CH3NH3+ and H2O
CH3NH3+ here is strong acid
CH3NH3+ formed = 7 mmol
Volume of Solution = 22.2222 + 40 = 62.2222 mL
Ka of CH3NH3+ = Kw/Kb = 1.0E-14/4.4E-4 = 2.273*10^-11
concentration ofCH3NH3+,c = 7 mmol/62.2222 mL = 0.1125 M
CH3NH3+ + H2O -----> CH3NH2 + H+
0.1125 0 0
0.1125-x x x
Ka = [H+][CH3NH2]/[CH3NH3+]
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 ((2.273*10^-11)*0.1125) = 1.599*10^-6
since c is much greater than x, our assumption is correct
so, x = 1.599*10^-6 M
[H+] = x = 1.599*10^-6 M
use:
pH = -log [H+]
= -log (1.599*10^-6)
= 5.7962
Answer: 5.80
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