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
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Species acting as both Bronsted acid and base is...Pure water is added into the following solutions causing a 10% increase in volume of each. The greatest % change in pH w...A volume of 20 ml of 0.8 M-HCN solution is mixed with 80 ml of 0.4 M-NaCN solution. Calculate the pH of the resulting so...The solubility products of MA, MB, MB and MD are 1.8 × 10-10, 4 × 10-3, 4 × 10-8 and 6 × 10-5 respec...Common hydrolysis so products of XeF2 & XeF4 are :-...