Question

1 mole of ch3nh2 is mixed with 0.08 mole hcl and diluted to 1litre, will it acts as a buffer solution

Answer 1

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The reaction is this:
CH3NH2 + H+ ---> CH3NH3^+
When 0.1 mole of CH3NH2 and 0.08
mole of HCl react, this is what remains
after the reaction:
0.02 mol CH3NH2
0.08 mol CH3NH3^+
Since we now have a buffer, we use the
Henderson-Hasselbalch Equation:
pH = pKa + log (base/acid)
Since we have the Kb, let us use it to get
the pKa:
pKb = - log 5*10^-4 = 3.30103
pKa = 14 - 3.30103 = 10.69897
Now, use the H-H Equation:
pH = 10.69897 + log (0.02 / 0.08) <---
since it's in 1 liter, these are the
molarities
pH = 10.69897 + (-0.60206)
pH = 10.09691
Now, we antilog the pH to get the H+
conc
[H+] = 10^-pH = 10^-10.09691 = 8 x
10^-11 M