Question

Calculate the pH when 30.0 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M CH3NH2 (Kb = 4.4 x 10-4).

Answer 1

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