Question 1

The pH of 1 N aqueous solutions of HCl , CH _3 COOH and HCOOH follows the order:

Accepted Answer

Stronger the acid, lower is the pH or weaker the acid, higher the pH . The strength of the given acid is in the order HCl HCOOH CH _3 COOH Hence, their pH is in the order CH _3 COOH HCOOH HCl

Question 2

20 mL of 0.1 M acetic acid is mixed with 50 mL of potassium acetate. K_ of acetic acid =1.8 10^ -5 at 27^ C . Calculate the concentration of potassium acetate if pH of mixture is 4.8 .

Accepted Answer

For an acidic buffer, pH = p K_ + [ Salt ] [ Acid ] (i) Let the concentration of potassium acetate solution =x M 20 mL of 0.1 M acetic acid =20 0.1 millimol =2 millimol 50 mL of x M potassium acetate =x 50 millimol aligned & =50 x milimol \ pH & =4.8 (Given) aligned Substitute all value in eq. (i), aligned 4.8 & =- ( 1.8+10^ -5 )+ 50 x 2 \ 25 x & =0.06 \ x & =0.04 M aligned

Question 3

Which of the following molecules acts as a Lewis acid?

Accepted Answer

According to Lewis concept "Acids are electron acceptor and bases are electron donor" Since, electron deficient compounds also have an ability to accept electrons, these are regarded as acids. Trimethylborane $( CH )_3$ B have incomplete octet thus, act as Lewis acid.

Question 4

Which one of the following molecular hydrides acts as a Lewis acid?

Accepted Answer

Boron in B _2 H _6 is electron deficient

Question 5

100 ~mL of 2 M of formic acid ( p K_a=3.74 ) is neutralise by NaOH , at the equivalence point pH is

Accepted Answer

Sodium formate is present at the equivalence point. It is the salt of weak acid + strong base. So, final solution will be basic in nature. As we know pH =7+ p K_a 2 + C 2 where, C is concentration of salt. Total volume of solution =100+100=200 ~mL Concentration of salt (C)=2 100 200 =1 M pH =7+ 3.74 2 + [1] 2 pH =7+1.87+0 pH =8.87

Question 6

What is [ NH _4^ + ]in a solution that is 0.02 M NH _3 0.01 M KOH ? [K_b ( NH _3 )=1.8 10^ -5 ]

Accepted Answer

NH _4 OH NH _4^ + + OH ^ - K_b= [ NH _4^ + ] [ OH ^ - ] [ NH _4 OH ] 1.8 10^ -5 = [ NH _4^ + ][0.01] [0.02] [ NH _4^ + ]=3.6 10^ -5 M

Question 7

The solubility of silver bromide in hypo solution involves the formation of:

Accepted Answer

Hypo solution is Na _ 2 ~S _ 2 O _ 3 solution which is used in photography for fixing films \& prints. Photographic emulsions are made of AgBr. After developing, the film is put into hypo solution. This forms soluble complex with Ag . Na _ 2 ~S _ 2 O _ 3 + Ag Br Ag _ 2 ~S _ 2 O _ 3 Na _ 2 ~S _ 2 O _ 3 Na _ 5 [ Ag ( S _ 2 O _ 3 )_ 3 ]

Question 8

A solution of NH _ 4 Cl and NH _ 3 has pH =8.0 . Which of the following hydroxides may be precipitated when this solution is mixed with equal volume of 0.2 M of metal ion.

Accepted Answer

pH =8, pOH =6 ; [ OH ^ - ]=10^ -6 M Ionic product of Fe ( OH )_ 2 =0.2 (1 10^ -6 )^ 2 =2 10^ -13 > K _ sp (=8.1 10^ -16 )

Ionic Equilibrium — Previous Year Questions & Solutions