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
Which of the following species is more soluble in water:...Which of the following when added to 1.0 L of 0.5 M HCl would result in maximum increase in pH :-...At what pH, is the solubility of Zn(OH)2 minimum? What is the minimum solubility?Zn(OH)2(s) $\rightleftharpoons$ Zn2+ (a...For weak electrolyte, AB, the degree of ionization would be (V = volume of solution having 1 mole of electrolyte and K i...How many grams of HCl should be dissolved in sufficient water to get 500 ml of an aqueous solution of pH, 2.0?...