C_5H_4NCOOH, nicotinic acid, has K_a = 1.40 Times 10^-5, 25.00 mL of 0.100 M nic
ID: 952333 • Letter: C
Question
C_5H_4NCOOH, nicotinic acid, has K_a = 1.40 Times 10^-5, 25.00 mL of 0.100 M nicotinic acid are titrated with 0.100 M NaOH. Calculate the pH of the acid before any base has been added. Calculate the volume of base needed to reach the equivalence point. Calculate the pH when 15.00% of the base needed to reach the equivalence point has been added. Calculate the pH at the equivalence point of the titration. Calculate the pH when 115% of the base needed to reach the equivalence point have been added.Explanation / Answer
a)
initially only weak acid is present
we know that
for weak acids
[H+] = sqrt ( Ka x C)
so
[H+] = sqrt ( 1.4 x 10-5 x 0.1)
[H+] = 1.183 x 10-3
now
pH = -log [H+]
so
pH = -log 1.183 x 10-3
pH = 2.927
b)
we know that
at equivalence point
Ma x Va = Mb x Vb
so
0.1 x 25 = 0.1 x Vb
Vb = 25
so
25 ml of base is need to get to equivalence point
c)
given 15 % of 25 ml
so
volume of NaOH added = 0.15 x 25 = 3.75 ml
we know that
moles = conc x volume (L)
so
moles of NaOH added = 0.1 x 3.75 x 10-3 = 0.375 x 10-3
moles of weak acid HA present = 0.1 x 25 x 10-3 = 2.5 x 10-3
now
the reaction is
HA + NaOH ---> NaA + H20
we can see that
moles of HA reacted = moles of NaoH added = 0.375 x 10-3
moles of NaA formed = moles of NaoH added = 0.375 x 10-3
so
finally
moles of HA remaining = 2.5 x 10-3 - 0.375 x 10-3 = 2.125 x 10-3
moles of NaA = 0.375 x 10-3
now
HA and NaA form a buffer solution
for buffers
pH = pKa + log [ salt / acid ]
also
pKa = -log Ka
so
pH = -log 1.4 x 10-5 + log [ 0.375 x 10-3 /2.125 x 10-3 ]
pH = 4.1
so
the pH is 4.1
d)
now
at equivalence point
moles of base added = moles of acid present = 0.1 x 25 x 10-3 = 2.5 x 10-3
now
the reaction is
HA + NaOH ---> NaA + H20
all the acid and base is converted to NaA
moles of NaA = 2.5 x 10-3
final volume = 25 + 25 = 50 ml
now
[NaA] = 2.5 x 10-3 x 1000 / 50 = 0.05
now
NaA is a weak base
for weak bases
[OH-] = sqrt( Kb x C)
also
Kb = Kw / Ka
so
[OH-] = sqrt ( 10-14 x 0.05 / 1.4 x 10-5]
[OH-] = 5.976 x 10-6
pOH = -log 5.976 x 10-6
pOH = 5.223
now
pH = 14 - pOH
so
pH = 14 - 5.223
pH = 8.777
e)
given
115 % of 25 ml
so
volume of NaOH added = 1.15 x 25 = 28.75
so
moles of NaOH added = 0.1 x 28.75 x 10-3 = 2.875 x 10-3
moles of acid present = 2.5 x 10-3
now
the reaction is
HA + NaOH ---> NaA + H20
moles of NaoH reacted = moles of HA = 2.5 x 10-3
so
moles of NaOH remaining = 2.875 x 10-3 - 2.5 x 10-3 = 0.375 x 10-3
now
final volume = 25 + 28.75 = 53.75 ml
[NaOH] = 0.375 x 10-3 x 1000 / 53.75 = 6.9767 x 10-3
now
NaOH ---> Na+ + OH-
so
[OH-] = [NaOH] = 6.9767 x 10-3
pOH = -log 6.9767 x 10-3
pOH = 2.156
now
pH = 14 - pOH
so
pH 14 - 2.156
pH = 11.844
so
pH of the solution is 11.844