1)
The reaction between HBr and CH3NH2 is
CH3NH2(aq) + H+(aq) -------> CH3NH3+(aq)
1:1 molar reaction
given moles of CH3NH2 = ( 0.400mol/1000ml)×30ml = 0.012mol
given moles of HBr = (0.200mol/1000ml) × 30ml = 0.0060mol
0.0060moles of HBr react with 0.0060moles of CH3NH2 to give 0.0060moles of CH3NH3+
After mixing
number of moles of CH3NH3+ = 0.0060mol
number of moles of CH3NH2 = 0.0120mol - 0.0060mol = 0.0060mol
Total volume = 60ml
[CH3NH3+] = (0.0060mol/60ml) × 1000ml = 0.10M
[CH3NH2] = (0.0060mol/60ml)× 1000ml = 0.10M
pKb= -logKa
pKb of CH3NH2= - log(4.4 ×10-4) = 3.36
pKa = 14 - pKb
pKa of CH3NH3+ = 14 - 3.36 = 10.64
Henderson - Hasselbalch equation is
pH = pKa + log([A-]/[HA])
pH = 10.64 + log ( 0.10M/0.10M)
pH = 10.64 + 0
pH = 10.64
2)
Moles of HBr = ( 0.200mol/1000ml) × 90ml = 0.018mol
moles of CH3NH2 = (0.400mol/1000ml ) × 30ml = 0.0120mol
0.0120moles of HBr react with 0.0120moles of CH3NH2
Remaining moles of HBr = 0.0060mol
Total volume = 120ml
[HBr] = ( 0.0060mol/120ml) × 1000ml = 0.050M
[H+] = 0.050M
pH = -log[H+]
pH = -log( 0.050M)
pH = 1.30
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