Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

a) For 230.0 mL of a buffer solution that is 0.205 M in HCHO2 and 0.275 M in KCHO2, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

b) For 230.0 mL of a buffer solution that is 0.320 M in CH3CH2NH2 and 0.290 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Answer 1

a)
Ka = 1.8 x 10^-4
pKa = 3.74
pH = 3.74 + log 0.275 / 0.205=3.86 ( initial pH of the buffer )

moles formic acid = 0.205 M x 0.230 L=0.0471 moles
moles formate = 0.275 M x 0.230 L =0.0632
the effect of the added 0.010 mol OH- would be to decrease the moles of formic acid by 0.010 and increases the moles of formate by 0.010 by the reaction
HCOOH + OH- = HCOO-
moles HCOOH = 0.0471 - 0.010 = 0.0371
moles HCOO- = 0.0632 + 0.010=0.0732
concentration HCOOH = 0.0371 / 0.230 L=0.161 M
concentration HCOO- = 0.0732/ 0.230 =0.318
pH = 3.74 + log 0.318 / 0.161= 4.03 ( final pH)


c)
Kb of ethylammine = 4.3 x 10^-4
pKb = 3.25
pOH = 3.37 + log 0.255 / 0.275 = 3.34
pH = 14 - 3.34 =10.7 ( initial pH)
moles CH3CH2NH3+ = 0.290 x 0.230 L=0.0667
moles CH3CH2NH2 = 0.320 x 0.230 L=0.0736
CH3CH2NH3+ + OH- = CH3CH2NH2 + H2O
moles CH3CH2NH3+ = 0.0667 - 0.010 =0.0567
moles CH3CH2NH2 = 0.0736 + 0.010 =0.0836
concentration CH3CH2NH3+ = 0.0567 / 0.230 =0.246 M
concentration CH3CH2NH2 = 0.0836 / 0.230=0.363 M
pOH = 3.37 + log 0.246/ 0.363=3.20
pH = 10.8