Question

For 220.0 mL of a buffer solution that is 0.2869 M in CH3CH2NH2 and 0.2624 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0200 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 {0.2624/0.2869}

= 3.213

use:

PH = 14 - pOH

= 14 - 3.213

= 10.787

Answer: 10.79

2)

mol of NaOH added = 0.02 mol

C3HCH2NH3+ will react with OH- to form CH3CH2NH2

Before Reaction:

mol of CH3CH2NH2 = 0.2869 M *0.22 L

mol of CH3CH2NH2 = 0.0631 mol

mol of C3HCH2NH3+ = 0.2624 M *0.22 L

mol of C3HCH2NH3+ = 0.0577 mol

after reaction,

mol of CH3CH2NH2 = mol present initially + mol added

mol of CH3CH2NH2 = (0.0631 + 0.02) mol

mol of CH3CH2NH2 = 0.0831 mol

mol of C3HCH2NH3+ = mol present initially - mol added

mol of C3HCH2NH3+ = (0.0577 - 0.02) mol

mol of C3HCH2NH3+ = 0.0377 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 {3.773*10^-2/8.312*10^-2}

= 2.909

use:

PH = 14 - pOH

= 14 - 2.9088

= 11.0912

Answer: 11.09