Question

Please find the pH and pOH for the 4 highlighted acids and 4 highlighted bases. Since Ka and Kb are given, we can basically just plug in the

-log, right? If not can you please show me? Thanks!!

Answer 1

1)

all the given acids are weak acids

noow

consider the weak acid to be HA

then

HA --> H+ + A-

Ka = [H+] [A-] / [HA]

using ICE table

initial concentration of HA, H+ , A- are 0.1 ,0 , 0

change in concentration of HA, H+ , A- are -y, +y , +y

equilibrium concentration of HA, H+ , A- are 0.1-y , y, y

now

Ka = [H+] [A-] /[HA]

Ka = [y] [y] / [0.1-y]

y2 = Ka (0.1-y)

y + (Ka x y) - 0.1Ka = 0

solve this equation to get the value of y

then

[H+] = y

now

pH = -log [H+]

a) for HCl02 , Ka = 1.2 x 10-2 = 0.012

so

y + 0.012y - 0.0012 = 0

y = 0.0291568

now

[H+] = y = 0.0291568

pH = -log [H+]

pH = -log 0.0291568

pH = 1.53526

so

pH of 0.1 M HCl02 is 1.53526

b) for HC2H2Cl02 , Ka = 1.35 x 10-3 = 0.00135

so

y + 0.00135y - 0.000135 = 0

y = 0.01096354

now

[H+] = y = 0.01096354

pH = -log [H+]

pH = -log 0.01096354

pH = 1.96

so

pH of 0.1 M HC2H2Cl02 is 1.96

c)

for HF , Ka = 7.2 x 10-4

so

y + (7.2 x 10-4)y - (7.2 x 10-5) = 0

y = 8.133 x 10-3

now

[H+] = y = 8.133 x 10-3

pH = -log [H+]

pH = -log 8.133 x 10-3

pH = 2.09

so

pH of 0.1 M HF is 2.09

d)

for HN02 , Ka = 4 x 10-4

so

y + ( 4 x10-4) y - ( 4 x 10-4) = 0

y = 6.1277 x 10-3

now

[H+] = y = 6.1277 x 10-3

pH = -log [H+]

pH = -log 6.1277 x 10-3

pH = 2.2127

so

pH of 0.1 M HN02 is 2.2127

2)

in case of CH3NH2 the reaction is

CH3NH2 + H20 --> CH3NH3+ + OH-

Kb = [CH3NH3+ ] [OH-] / [CH3NH2]

using ICE table

initial concentration of CH3NH2, CH3NH3+ , OH- are 0.1 ,0 , 0

change in concentration of CH3NH2, CH3NH3+ , OH- are -y, +y , +y

equilibrium concentration of CH3NH2 , CH3NH3+ , OH- are 0.1-y , y, y

now

Kb = [CH3NH3+] [OH-] /[CH3NH2]

Kb = [y] [y] / [0.1-y]

y2 = Kb (0.1-y)

y + (Kb x y) - 0.1Kb = 0

solve this equation to get the value of y

then

[OH-] = y

now

pOH = -log [OH-]

a) for 0.1 M CH3NH2

y2 + ( 4.38 x 10-4)y - (4.38 x 10-5) = 0

y = 6.40278 x 10-3

now

[OH-] = 6.40278 x 10-3

pOH = -log [OH-]

pOH = -log 6.40278 x 10-3

pOH = 2.1936

b)

for 0.1 M C2H5NH2

y2 + ( 5.6 x 10-4)y - (5.6 x 10-5) = 0

y = 7.20855 x 10-3

now

[OH-] = 7.20855 x 10-3

pOH = -log [OH-]

pOH = -log 7.20855 x 10-3

pOH = 2.142

c) for 0.1 M C6H5NH2

y2 + ( 3.8 x 10-10)y - (3.8 x 10-11) = 0

y = 6.1644 x 10-6

now

[OH-] = 6.1644 x 10-6

pOH = -log [OH-]

pOH = -log 6.1644 x 10-6

pOH = 5.21

d) for 0.1 M C5H5N

y2 + ( 1.7 x 10-9)y - (1.7 x 10-10) = 0

y = 1.30384 x 10-5

now

[OH-] = 1.30384 x 10-5

pOH = -log [OH-]

pOH = -log 1.30384 x 10-5

pOH = 4.88477