Question

# 1. Calculate the pH of a solution that is 0.050 M acetic acid and 0.010 M...

1. Calculate the pH of a solution that is 0.050 M acetic acid and 0.010 M phenylacetic acid.

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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