Engineering Chemistry S Datta

196 ENGINEERING CHEMISTRY
∴ x 2 × 1.0 × 10 –8 x – 10 –14 = 0
∴ x = 9.5 × 10 –8 (M)
∴ [OH – ] = 9.5 × 10 –8 g. ion l –1
∴ pOH = 8 – log 9.5 = 6.98
∴ pH = 14 – 6.98 = 7.02.
Example 4. (a) Calculate the molar concentration of an CH 3
COOH solution. (Given
K a
= 1.8 × 10 –5 . α = 0.02)
(b) Calculate the degree of dissociation of 0.01 (M) AcOH.
Sol. CH 3
COOH H + + CH 3
COO –
at equilibrium: C(1 – α) Cα Cα
+ −
[H ] [CH3
COO ]
∴ K a
=
[CH3
COOH]
0.02C × 0.2C 0.0004 × C
= =
C( 1 − 0.02)
0.98
−5
18 . × 10 × 098 .
∴ c = molar
0.
004
= 0.044 molar.
(b) α =
K
18 . × 10−5
=
C 001 .
= 4.24 × 10 –2 .
Example 5. A buffer solution contains 0.1 mole CH 3
COOH and 0.1 mole of CH 3
COONa
per litre. Calculate the pH of buffer solution. (pK a
= 4.74).
Sol. According to Henderson equation pH of a buffer solution
pH = pK a
+ log [ salt ]
[acid]
= 4.74 + log 01 .
= 4.74.
01 .
Example 6. pH of blood is 7.4. Calculate the ratio of HCO
–
3
and H 2
CO 3
. Given K a
of
H 2
CO 3
= 4.5 × 10 –7 .
Sol. H 2
CO 3
H + + HCO – 3
,
∴
+ −
[H ] [HCO 3 ]
K a
=
[H2CO 3]
∴
[HCO − ] K
3
4.
5 10−7 4.
5 10−7
= a ×
×
=
=
[H CO ] [H + ] [H + ] 398 . × 10−8
2 3
~_ 11
pH = 7.4
∴ – log 10
[H + ] = 7.4 ∴ [H + ] = 10 –7.4 = 3.98 × 10 –8 g.ion l –1
Example 7. The solubility product of CaF 2
in water at 18°C is 3.45 × 10 –11 . Calculate the
solubility of CaF 2
in water.
Sol. K sp
= [Ca +2 ] [F – ] 2
= [S] × [2S] 2 where, S = solubility.
= 4S 3