HARD
JEE Main
IMPORTANT
Earn 100

The pKa of acetyl salicylic acid (aspirin) is 3·5 . The pH of gastric juice in the human stomach is about 2-3 and the pH in the small intestine is about 8. Aspirin will be

34.85% studentsanswered this correctly

Important Questions on Equilibrium

HARD
JEE Main
IMPORTANT

An inert gas is added to the following equilibrium,

A(s)+2B(g)3C(g) at constant pressure. The equilibrium 

HARD
JEE Main
IMPORTANT

Calculate the molar solubility of  AgCl in 1.0 M NH3.Ksp(AgCl)=1.8×10-10, Kf AgNH32+=1.7×107.

Give your answer up to two places of decimal.

HARD
JEE Main
IMPORTANT

Calculate the pH at the equivalence point, when a solution of 0.1 M CH3COOH is titrated with a solution of 0.1 M NaOH, KaCH3COOH=1.8×10-5.

(Choose your answer from these values : 4.74/8.72/11.74/7.12)

HARD
JEE Main
IMPORTANT

The value of Kp for the reaction PCl5PCl3+Cl2 is 1.78 at 250°C. The fraction of dissociation is a × 10-4 at equilibrium when 0.40 mole of PCl5 is vaporised in a vessel containing 0.20 mole of Cl2 gas when the volume is kept constant at 4 litres.

Calculate the value of a.

 

HARD
JEE Main
IMPORTANT
Ka for butyric acid is 2.0×10-5. Calculate the pH of 0.2 M aqueous solution of sodium butyrate.
HARD
JEE Main
IMPORTANT
How many moles of NaOH can be added to one litre of a solution of 0.1 M in NH3 and 0.1 M in NH4Cl, without changing the pOH more than one unit? Assume no change in volume and pKb=4.75.
HARD
JEE Main
IMPORTANT

Calculate the solubility of Mg(OH)2 in 0.05 M NaOH.

KspMg(OH)2=8.9×10-12.

(Report your answer in multiples of 10-9.)

HARD
JEE Main
IMPORTANT

0.1 mole of each ethyl alcohol and acetic acid are allowed to react, and at equilibrium, the acid is exactly neutralised by 100 mL of 0.85 N NaOH. If no hydrolysis of an ester is supposed to have been undergone, the Equilibrium constant, which is given as a × 10-3, give the value of a.

Give the answer to the nearest integer value.