What is the pH after 70 mL of 0.20 M HCI has been added to 50 mL 0.60 M H3CNH2? H3CNH2 has a pKb =3.42.

Question

Please help with this question, I'm not sure how to go about it

Answer 1

Adding HCl (a strong acid) to CH₃NH₂ (methylamine, a weak base) forms a buffer. This buffer consists of

CH₃NH₂ and CH₃NH₃⁺, i.e. a weak base and the salt (conjugate acid) of that weak base. The pH of a buffer can be calculated using the Henderson Hasselbalch equation pOH = pKb + log [salt]/[base]. We know the pKb (it's given) and we can calculate log [salt]/[base].

CH₃NH₂ + HCl ==> CH₃NH₃⁺ + Cl^-

moles CH₃NH₂ = 0.050 L x 0.60 mol/L = 0.030 CH₃NH₂ initially present

moles HCl added = 0.070 L x 0.20 mol/L = 0.014 moles moles HCl added

moles CH₃NH₂ used up = 0.014

moles CH₃NH₂ remaining after HCl = 0.030 - 0.014 = 0.016 moles CH₃NH₂ final

moles CH₂NH₃⁺ formed by addition of HCl = 0.014 moles CH₃NH₃⁺ final

Final volume = 70 ml + 50 ml = 120 ml = 0.12 L

Final [CH₃NH₂] = 0.016 mol/0.12 L = 0.133 M

Final [CH₃NH₃⁺] = 0.014 mol/0.12 L = 0.117 M

pOH = pKb + log [salt]/[base]

pOH = 3.42 + log (0.117/0.133) = 3.42 + (-0.057)

pOH = 3.36

pH = 10.6