Question

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

For 300.0mL of pure water, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

For 300.0mL of a buffer solution that is 0.240M in HCHO2 and 0.280M in KCHO2, calculate the initial pH and the final pHafter adding 0.010 mol of NaOH.

For 300.0mL of a buffer solution that is 0.305M in CH3CH2NH2 and 0.285M in CH3CH2NH3Cl, calculate the initial pHand 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/ 0.300 L=0.0333 M
pOH = - log (0.0333)= 1.47
pH = 14 - pOH = 14 - 1.47=12.53


b)
Ka = 1.8 x 10^-4
pKa = 3.74
pH = 3.74 + log 0.280 / 0.240 =3.80 ( initial pH of the buffer )

moles formic acid = 0.240 M x 0.300 L=0.072
moles formate = 0.280 M x 0.300 L =0.084

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 = 0.072 - 0.010 = 0.062
moles HCOO- = 0.084 + 0.010=0.094

concentration HCOOH = 0.062 / 0.300 L=0.206 M
concentration HCOO- = 0.094 / 0.300 =0.313
pH = 3.74 + log 0.313 / 0.206 = 3.92

pH=3.92 (final pH)

c)
Kb of ethylammine = 4.3 x 10^-4
pKb = 3.37
pOH = 3.37 + log 0.285 / 0.305 = 3.34
pH = 14 - 3.34 =10.7 ( initial pH)

moles CH3CH2NH3+ = 0.285 x 0.300 L=0.0855
moles CH3CH2NH2 = 0.305 x 0.300 L=0.0915

CH3CH2NH3+ + OH- = CH3CH2NH2 + H2O

moles CH3CH2NH3+ = 0.0855 - 0.010 =0.0755
moles CH3CH2NH2 = 0.0915 + 0.010 =0.1015

concentration CH3CH2NH3+ = 0.0755 / 0.300 =0.251 M
concentration CH3CH2NH2 = 0.1015 / 0.300=0.338 M

pOH = 3.37 + log 0.251 / 0.338=3.24

pH = 10.76 (final pH)