Ionic EquilibriumHard
Question
A weak acid (HA) after treatment with 12 mL of 0.1 M strong base (BOH) has a pH of 5. At the end point, the volume of same base required is 27 mL. Ka of acid is (log2 = 0.3)
Options
A.1.8 × 10-5
B.8 × 10-6
C.1.8 × 10-6
D.8 × 10-5
Solution
m. moles of HA taken = 27 × 0.1 = 2.7
HA + OH- A- + H2O
t=0 2.7 1.2
teq 1.5 - 1.2
pH = pKa + log
⇒ 5 = pKa + log 
= pKa + log
∴ pKa = 5.1 ⇒ Ka = 8 × 10-6.
HA + OH- A- + H2O
t=0 2.7 1.2
teq 1.5 - 1.2
pH = pKa + log
⇒ 5 = pKa + log = pKa + log∴ pKa = 5.1 ⇒ Ka = 8 × 10-6.
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
As acidic buffer solution can be prepared by mixing the solutions of :...The reverse process of neutralisation is:...The active ingredient in aspirin is acetyl salicylic acidwith Ka = 4.0 × 10–9. The pH of the solution obtained by dissol...How many litres of water must be added to 1litre of an aqueous solution of HCl with a pH of 1 litre to create an aqueous...Find the number of P - O - P linkage in (HPO3)n where n = 3...