Question

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

b. For 210.0mL of a buffer solution that is 0.210M in HCHO2 and 0.295M in KCHO2, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

c. For 210.0mL of a buffer solution that is 0.265M in CH3CH2NH2 and 0.240M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Answer 1

a)
pure water : pH = 7.00
moles NaOH = 0.010
[OH-]= 0.010 mol/ 0.210 L=0.0476 M
pOH = - log 0.0476= 1.32
pH = 14 - pOH = 14 - 1.32=12.67


b)
Ka = 1.8 x 10^-4
pKa = 3.74
pH = 3.74 + log 0.295 / 0.210=3.88 ( initial pH of the buffer )

moles formic acid = 0.210 M x 0.210 L=0.0441
moles formate = 0.295 M x 0.210 L =0.06195
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 = 0.0441 - 0.010 = 0.0341
moles HCOO- = 0.06195 + 0.010=0.0519
concentration HCOOH = 0.0341 / 0.210 L=0.162 M
concentration HCOO- = 0.0519 / 0.210 =0.247
pH = 3.74 + log 0.247 / 0.162 = 3.92 ( final pH)


c)
Kb of ethylammine = 4.3 x 10^-4
pKb = 3.37
pOH = 3.37 + log 0.240 / 0.265 = 3.33
pH = 14 - 3.33 =10.67 ( initial pH)
moles CH3CH2NH3+ = 0.240 x 0.210 L=0.0504
moles CH3CH2NH2 = 0.265 x 0.210 L=0.0556
CH3CH2NH3+ + OH- = CH3CH2NH2 + H2O
moles CH3CH2NH3+ = 0.0504 - 0.010 =0.0404
moles CH3CH2NH2 = 0.0556 + 0.010 =0.0656
concentration CH3CH2NH3+ = 0.0404 / 0.210 =0.192 M
concentration CH3CH2NH2 = 0.0656 / 0.210=0.312 M
pOH = 3.37 + log 0.192 / 0.312=3.16
pH = 10.8