Question

Calculate the pH at each of the following points along the titration of 25.0 mL of 0.185 M HNO2 with 0.185 M NaOH. [K, = 4.3x10-61 pka = -109(4.3x10-6 - 5136 5.36 +log 185/. 185 = 5. 4 5 20 0 5.0 odstoty D oonboe 3.1 HOMO 4 5: 2 . SI. OLD 0 mL of NaOH added pH = _ 8.5 mL of NaOH added pH = 12.5 mL of NaOH added pH = 22.0 mL of NaOH added pH =_ 25.0 mL of NaOH added pH = 30.0 mL of NaOH added pH = _

Answer 1

a)when 0.0 mL of NaOH is added

HNO2 dissociates as:

HNO2 -----> H+ + NO2-

0.185 0 0

0.185-x x x

Ka = [H+][NO2-]/[HNO2]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((4.3*10^-6)*0.185) = 8.919*10^-4

since c is much greater than x, our assumption is correct

so, x = 8.919*10^-4 M

use:

pH = -log [H+]

= -log (8.919*10^-4)

= 3.0497

Answer: 3.05

b)when 8.5 mL of NaOH is added

Given:

M(HNO2) = 0.185 M

V(HNO2) = 25 mL

M(NaOH) = 0.185 M

V(NaOH) = 8.5 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.185 M * 25 mL = 4.625 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.185 M * 8.5 mL = 1.5725 mmol

We have:

mol(HNO2) = 4.625 mmol

mol(NaOH) = 1.5725 mmol

1.5725 mmol of both will react

excess HNO2 remaining = 3.0525 mmol

Volume of Solution = 25 + 8.5 = 33.5 mL

[HNO2] = 3.0525 mmol/33.5 mL = 0.0911M

[NO2-] = 1.5725/33.5 = 0.0469M

They form acidic buffer

acid is HNO2

conjugate base is NO2-

Ka = 4.3*10^-6

pKa = - log (Ka)

= - log(4.3*10^-6)

= 5.367

use:

pH = pKa + log {[conjugate base]/[acid]}

= 5.367+ log {4.694*10^-2/9.112*10^-2}

= 5.078

Answer: 5.08

c)when 12.5 mL of NaOH is added

Given:

M(HNO2) = 0.185 M

V(HNO2) = 25 mL

M(NaOH) = 0.185 M

V(NaOH) = 12.5 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.185 M * 25 mL = 4.625 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.185 M * 12.5 mL = 2.3125 mmol

We have:

mol(HNO2) = 4.625 mmol

mol(NaOH) = 2.3125 mmol

2.3125 mmol of both will react

excess HNO2 remaining = 2.3125 mmol

Volume of Solution = 25 + 12.5 = 37.5 mL

[HNO2] = 2.3125 mmol/37.5 mL = 0.0617M

[NO2-] = 2.3125/37.5 = 0.0617M

They form acidic buffer

acid is HNO2

conjugate base is NO2-

Ka = 4.3*10^-6

pKa = - log (Ka)

= - log(4.3*10^-6)

= 5.367

use:

pH = pKa + log {[conjugate base]/[acid]}

= 5.367+ log {6.167*10^-2/6.167*10^-2}

= 5.367

Answer: 5.37

d)when 22.0 mL of NaOH is added

Given:

M(HNO2) = 0.185 M

V(HNO2) = 25 mL

M(NaOH) = 0.185 M

V(NaOH) = 22 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.185 M * 25 mL = 4.625 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.185 M * 22 mL = 4.07 mmol

We have:

mol(HNO2) = 4.625 mmol

mol(NaOH) = 4.07 mmol

4.07 mmol of both will react

excess HNO2 remaining = 0.555 mmol

Volume of Solution = 25 + 22 = 47 mL

[HNO2] = 0.555 mmol/47 mL = 0.0118M

[NO2-] = 4.07/47 = 0.0866M

They form acidic buffer

acid is HNO2

conjugate base is NO2-

Ka = 4.3*10^-6

pKa = - log (Ka)

= - log(4.3*10^-6)

= 5.367

use:

pH = pKa + log {[conjugate base]/[acid]}

= 5.367+ log {8.66*10^-2/1.181*10^-2}

= 6.232

Answer: 6.23

e)when 25.0 mL of NaOH is added

Given:

M(HNO2) = 0.185 M

V(HNO2) = 25 mL

M(NaOH) = 0.185 M

V(NaOH) = 25 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.185 M * 25 mL = 4.625 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.185 M * 25 mL = 4.625 mmol

We have:

mol(HNO2) = 4.625 mmol

mol(NaOH) = 4.625 mmol

4.625 mmol of both will react to form NO2- and H2O

NO2- here is strong base

NO2- formed = 4.625 mmol

Volume of Solution = 25 + 25 = 50 mL

Kb of NO2- = Kw/Ka = 1*10^-14/4.3*10^-6 = 2.326*10^-9

concentration ofNO2-,c = 4.625 mmol/50 mL = 0.0925M

NO2- dissociates as

NO2- + H2O -----> HNO2 + OH-

0.0925 0 0

0.0925-x x x

Kb = [HNO2][OH-]/[NO2-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.326*10^-9)*9.25*10^-2) = 1.467*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.467*10^-5 M

[OH-] = x = 1.467*10^-5 M

use:

pOH = -log [OH-]

= -log (1.467*10^-5)

= 4.8337

use:

PH = 14 - pOH

= 14 - 4.8337

= 9.1663

Answer: 9.17

6)when 30.0 mL of NaOH is added

Given:

M(HNO2) = 0.185 M

V(HNO2) = 25 mL

M(NaOH) = 0.185 M

V(NaOH) = 30 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.185 M * 25 mL = 4.625 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.185 M * 30 mL = 5.55 mmol

We have:

mol(HNO2) = 4.625 mmol

mol(NaOH) = 5.55 mmol

4.625 mmol of both will react

excess NaOH remaining = 0.925 mmol

Volume of Solution = 25 + 30 = 55 mL

[OH-] = 0.925 mmol/55 mL = 0.0168 M

use:

pOH = -log [OH-]

= -log (1.682*10^-2)

= 1.7742

use:

PH = 14 - pOH

= 14 - 1.7742

= 12.2258

Answer: 12.23