Ionic EquilibriumHard
Question
If the equilibrium constant for the reaction of weak acid HA with strong base is 109, then pH of 0.1 M Na A is
Options
A.5
B.9
C.7
D.8
Solution
HA + NaOH → NaA + H2O; or HA + OH- → A- + H2O
Keq. = 109 =
Also HA ⇋ H+ + A- Ka = 
∴
or Ka = 109 × 10-14 = 10-5 Thus for
A- + H2O ⇋ HA + OH
[OH-] = Ch = C
= 10-5 M
∴ [H+]= 10- and pH = 9
Keq. = 109 =
Also HA ⇋ H+ + A- Ka = ∴
or Ka = 109 × 10-14 = 10-5 Thus for A- + H2O ⇋ HA + OH
[OH-] = Ch = C
= 10-5 M∴ [H+]= 10- and pH = 9
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