Question

What is the pH of 50 mL of a 1.0 mol/L methylamine (CH3NH2) solution after the addition of 50 mL of a 0.5 mol/L solution of HCl?

5. What is the pH of 50 mL of a 1.0 mol/L methylamine (CH3NH2) solution after the addition of 50 mL of a 0.5 mol/L solution of HCI?

Answer 1

Number of moles of methyl amine = molarity of methyl amine solution X volume of methyl amine solution in L= 1 M X (50/1000) L =0.05 mole

Number of moles of HCl = molarity of HCl X volume of HCl solution in L= 0.5 M X (50/1000) L =0.025 mole

                       CH3NH2 + HCl $\rightleftharpoons$ CH3NH3⁺ + Cl^-

Initial moles       0.05      0.025           0              0

Change moles    -0.025    - 0.025      +0.025      +0.025

Equlibrium miles   0.025       0             0.025         0.025

Concentration of CH3NH2 at equilibrium = number of moles / total volume in L

                                                                     = 0.025 mole/0.1 L = 0.25 M

Concentration of CH3NH3⁺ at equilibrium = number of moles / total volume in L

                                                                     = 0.025 mole/0.1 L = 0.25 M

For given buffer solution

pOH = pKb(CH3NH2) + log [CH3NH3⁺]/ [CH3NH2]

pOH =3.36 + log (0.25/0.25)

pOH = 3.36

pH + pOH =14

pH= 10.64 Answer