Question

Calculate the pH of a buffer solution that is 1.25 M CH₃NH₂ and 1.00 M CH₃NH₃Cl. K_b = 8.85 x 10^-4 and what would happen after 35.0 mL of 0.45 M HNO₃ is added to 250.0 mL of the buffer solution, what would the pH of the solution be after the HNO₃(aq) is added?

Answer 1

initially

pOH = pKb + log [CH3NH3Cl] / [CH3NH2]

pKb = - log Kb = - log [8.85 x 10^-4] = 3.05

pOH = 3.05 + log [1.00] / [1.25]

pOH = 3.05 - 0.097

pOH = 2.95

pH = 14 - 2.95

pH = 11.05

now after HNO3 added

initially

millimoles of CH3NH2 = 250 x 1.25 = 312.5

millimoles of CH3NH3Cl = 250 x 0.1 = 250

millimoles of HNO3 added = 35 x 0.45 = 15.75

after HNO3 added

millimoles of CH3NH2 = 312.5 - 15.75 = 296.75

millimoles of CH3NH3⁺ = 250 + 15.75 = 265.75

total volume = 250 + 35 = 285 mL

[CH3NH2] = 296.75 / 285 = 1.04 M

[CH3NH3⁺] = 265.75 / 285 = 0.932 M

pOH = 3.05 + log [0.932] / [1.04]

pOH = 3.05 - 0.047

pOH = 3.00

pH = 14 - 3.00

pH = 11.0