Question

For 220.0 mL of a buffer solution that is 0.3028 M in CH₃CH₂NH₂ and 0.2783 M in CH₃CH₂NH₃Cl, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH(Kb=5.6⋅10⁻⁴).

Answer 1

1)
Kb = 5.6*10^-4

pKb = - log (Kb)
= - log(5.6*10^-4)
= 3.252

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.252+ log {0.2783/0.3028}
= 3.22

use:
PH = 14 - pOH
= 14 - 3.22
= 10.78
Answer: 10.78

2)

mol of NaOH added = 0.01 mol

CH3CH2NH3+ will react with OH- to form CH3CH2NH2

Before Reaction:
mol of CH3CH2NH2 = 0.3028 M *0.22 L
mol of CH3CH2NH2 = 0.0666 mol

mol of CH3CH2NH3+ = 0.2783 M *0.22 L
mol of CH3CH2NH3+ = 0.0612 mol

after reaction,
mol of CH3CH2NH2 = mol present initially + mol added
mol of CH3CH2NH2 = (0.0666 + 0.01) mol
mol of CH3CH2NH2 = 0.0766 mol

mol of CH3CH2NH3+ = mol present initially - mol added
mol of CH3CH2NH3+ = (0.0612 - 0.01) mol
mol of CH3CH2NH3+ = 0.0512 mol
since volume is both in numerator and denominator, we can use mol instead of concentration
Kb = 5.6*10^-4

pKb = - log (Kb)
= - log(5.6*10^-4)
= 3.252

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.252+ log {5.123*10^-2/7.662*10^-2}
= 3.077

use:
PH = 14 - pOH
= 14 - 3.077
= 10.923
Answer: 10.92