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

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.

Solutions

Expert Solution

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

queen_honey_blossom answered 1 year ago

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