1. Calculate the pH of a solution that is 0.050 M acetic acid and 0.010 M phenylacetic acid.
Express your answer using two decimal places.
2. Calculate the pH at the points in the titration of 25.00 mL of 0.136M HNO2 when the following amounts of 0.122M NaOH have been added. For HNO2, Ka=7.2
1) the pKa values are
for acetic acid Ka = 1.74 x 10-5
for phenylacetic acid = 4.9 x 10-5
for weak acids
[H+] = sqrt( Ka x C)
so for acetic acid
[H+] = sqrt ( 1.74 x 10-5 x 0.05)
[H+] = 9.32 x 10-4
for phenyl acetic acid
[H+] = sqrt ( 4.9 x 10-5 x 0.01 )
[H+] = 7 x 10-4
now
total [H+] = (9.32 x 10-4 )+ ( 7 x 10-4 )
total [H+] = 16.32 x 10-4 M
so
pH = -log [H+]
pH = -log 16.32 x 10-4
pH = 2.79
so the pH of the solution is 2.79
2)
moles = molarity x volume (L)
so
moles of HN02 taken = 0.136 x 25 x 10-3
moles of HN02 taken = 3.4 x 10-3
A) 10 ml of 0.122 M NaOH is added
moles of NaOH added = 0.122 x 10 x 10-3 = 1.22 x 10-3
the reaction given is
NaOH + HN02 ---> NaN02 + H20
moles of HN02 reacted = moles of NaOH added = 1.22 x 10-3
moles of HN02 remaining = 3.4 x 10-3 - 1.22 x 10-3
moles of Hn02 remaining = 2.18 x 10-3
also
moles of NaN02 formed = moles of HN02 reacted = 1.22 x 10-3
now
the combination of HN02 and NaN02 forms an acidic buffer
for acidic buffers
pH = pKa + log [ conjugate base/ acid]
so
for the given combination
pH = -logKa + log [ NaN02 / HN02]
we know that
molarity = moles / volume
as the final volume is same for both , they cancel out
so
pH = -log 7.2 x 10-4 + log [ 1.22 x 10-3 / 2.18 x 10-3 ]
pH = 2.89
So the pH is 2.89
B) 20 ml of 0.122 M NaOH is added
moles of NaOH added = 0.122 x 20 x 10-3 = 2.44 x 10-3
the reaction given is
NaOH + HN02 ---> NaN02 + H20
moles of HN02 reacted = moles of NaOH added = 2.44 x 10-3
moles of HN02 remaining = 3.4 x 10-3 - 2.44 x 10-3
moles of Hn02 remaining = 9.6 x 10-4
also
moles of NaN02 formed = moles of HN02 reacted = 2.44 x 10-3
now
the combination of HN02 and NaN02 forms an acidic buffer
for acidic buffers
pH = pKa + log [ conjugate base/ acid]
so
for the given combination
pH = -logKa + log [ NaN02 / HN02]
we know that
molarity = moles / volume
as the final volume is same for both , they cancel out
so
pH = -log 7.2 x 10-4 + log [ 2.44 x 10-3 / 9.6 x 10-4 ]
pH = 3.55
So the pH is 3.55
3) given
pH of HCl = 2.3
so
pH = -log [H+]
2.3 = -log [H+]
[H+] = 5 x 10-3 M
now
HCl ----> H+ + Cl-
as HCl is a very strong acid , it dissociates 100%
so
[H+] = [HCl] = 5 x 10-3 M
now
for NaOH given pH is 11
pOH = 14 - pH
pOH = 14 -11
pOH = 3
we know that
pOH = -log [OH-]
3 = -log [OH-]
[OH-] = 10-3
now
NaOH ---> Na+ + OH-
as NaOH is a very strong base
[OH-] = [NaOH] = 10-3 M
now
moles = molarity x volume (L)
so
moles of HCl = 5 x 10-3 x 0.1
moles of HCl = 5 x 10-4
moles of NaOH = 10-3 x 0.1
moles of NaOH = 1 x 10-4
now the reaction is
NaOH + HCl ---> NaCl + H20
from the above reaction
moles of HCl reacted = moles of NaOH added = 1 x 10-4
so
moles of HCl remaining = 5 x 10-4 - 1 x 10-4
moles of HCl remaining = 4 x 10-4
total volume = 100 + 100 = 200 ml = 0.2 L
so
molarity of HCl = moles / volume (L)
molarity of HCl = 4 x 10-4 / 0.2
molarity of HCl = 2 x 10-3 M
so
[H+] = [HCl] = 2 x 10-3
pH = -log 2 x 10-3
pH = 2.70
so the pH of the resulting solution is 2.70
4)
we know that
moles = molarity x volume (L)
so
moles of H2S04 taken = 1.7 x 10-2 x 50 x 10-3
moles of H2S04 taken = 8.5 x 10-4
also
moles of NaOH = 3.85 x 10-2 x 50 x 10-3
moles of NaOH = 1.925 x 10-3
the reaction between NaOH and H2S04 is
2NaOH + H2S04 ----> Na2S04 + 2H20
so
moles of NaOH reacted = 2 x moles of H2S04
so
moles of NaOH reacted = 2 x 8.5 x 10-4
moles of NaOH reacted = 1.7 x 10-3
so
moles of NaOH remaining = 1.925 x 10-3 - 1.7 x 10-3
moles of NaOH remaining = 2.25 x 10-4
total volume = 50 + 50 = 100 ml
total volume = 0.1 L
so
molarity of NaOH = moles / volume (L)
molarity of NaOH = 2.25 x 10-4 / 0.1
molarity of NaOH = 2.25 x 10-3
now
NaOH ---> Na+ + OH-
as NaOH is a strong base ,it dissociates 100%
so
[OH-] = [NaOH]
[OH-] = 2.25 x 10-3
we know that
poH = -log [OH-]
pOH = -log 2.25 x 10-3
pOH = 2.6478
now
pH = 14 - pOH
pH = 14 - 2.6478
pH = 11.35
so the pH of the resulting solution is 11.35
5)
the combination of C6H5COOH and NaC6H5C00 forms a acidic
buffer
for an acidic buffer
pH = pKa + log [conjugate base / acid ]
also
pKa = -log Ka
so
pH= -logKa + log [ NaC6H5COO/ C6H5COOH]
so
using given values
pH = -log 6.3 x 10-5 + log [ 3.5 x 10-2 / 1 x 10-2
]
pH = 4.74
so the pH of the given buffer is 4.74
6) the combination of NH3 and NH4Cl forms a basic buffer
we know that
for a basic buffer
pOH = pKb + log [conjugate acid / base]
for the given combination
pOH = pKb + log [NH4Cl/NH3]
also
pKb for Nh3 is 4.75
so
using given values
we get
pOH = 4.75 + log [0.151 / 0.408]
pOH = 4.32
now
pH = 14 - pOH
pH = 14 - 4.32
pH = 9.68
so the pH of the given buffer is 9.68
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