Page 265 - Theory and Problems of BEGINNING CHEMISTRY

254 ACID-BASE THEORY [CHAP. 17

ACID-BASE EQUILIBRIUM
+
17.8. The [H 3 O ] in 0.150 M benzoic acid (HC 7 H 5 O 2 )is3.08 × 10 −3 M. Calculate K a .
−→
+
Ans. HC 7 H 5 O 2 + H 2 O ←− H 3 O + C 7 H 5 O 2 −
[H 3 O ] = [C 7 H 5 O 2 ] = 3.08 × 10 −3
+
−
−3
[HC 7 H 5 O 2 ] = 0.150 − (3.08 × 10 ) = 0.147
−3 2
[H 3 O ][C 7 H 5 O 2 ] (3.08 × 10 )
+
−
K a = = = 6.45 × 10 −5
[HC 7 H 5 O 2 ] 0.147
−4
17.9. For formic acid, HCHO 2 , K a is 1.80 × 10 . Calculate [H 3 O ] in 0.0150 M HCHO 2 .
+
−→
Ans. HCHO 2 + H 2 O ←− H 3 O + CHO 2 −
+
Initial 0.0150 0 0
Change −x x x
Equilibrium 0.0150 − x x x
If we neglect x when it is subtracted from 0.0150, then
[H 3 O ][CHO 2 ] (x)(x) −4
−
+
K a = = = 1.80 × 10
[HCHO 2 ] 0.0150
2
x = 2.70 × 10 −6
x = 1.64 × 10 −3
This value of x is 10.9% of the concentration from which it was subtracted. The approximation is wrong.
The quadratic formula must be used:

(x)(x)
K a = = 1.80 × 10 −4
0.0150 − x
2
−4
x =−1.80 × 10 x + 2.70 × 10 −6
−4
2
x + 1.80 × 10 x − 2.70 × 10 −6 = 0
√
2
−b ± b − 4ac
x =
2a
−4 2
−6
−1.80 × 10 −4 + (1.80 × 10 ) − 4(−2.70 × 10 )
=
2
+
x = 1.56 × 10 −3 = [H 3 O ]
−3 2
(1.56 × 10 )
Check: = 1.8 × 10 −4
−3
0.0150 − (1.56 × 10 )
−
17.10. What is the CN concentration in 0.030 M HCN? K a = 6.2 × 10 −10 .
Ans. HCN + H 2 O ←− CN + H 3 O +
−→
−
Initial 0.030 0 0
Change −x x x
Equilibrium 0.030 − x x x
[CN ][H 3 O ] x 2 −10
−
+
K a = = = 6.2 × 10
[HCN] 0.030
−
x = 4.3 × 10 −6 = [CN ]