Q 1 :    

Phosphoric acid ionizes in three steps with their ionization constant values Ka1, Ka2 and Ka3 respectively, while K is the overall ionization constant.

Which of the following statements are true?

A. logK=logKa1+logKa2+logKa3
B. H3PO4 is a stronger acid than H2PO4- and HPO42-
C. Ka1>Ka2>Ka3
D. Ka1=Ka3+Ka22

Choose the correct answer from the options given below:                    [2025]

  • B, C and D only

     

  • A, B and C only

     

  • A and B only

     

  • A and C only

     

(2)

K=Ka1×Ka2×Ka3

Hence, logK=logKa1+logKa2+logKa3

It is difficult to remove a proton from negatively charged ion, hence, H3PO4 is stronger acid than H2PO4- and HPO42-.

Higher order ionisation constants (Ka2,Ka3) are smaller than the lower order ionisation constant (Ka1).  Hence, Ka1>Ka2>Ka3.



Q 2 :    

For a weak acid HA, the percentage of dissociation is nearly 1% at equilibrium. If the concentration of acid is 0.1 mol L-1, then the correct option for its Ka at the same temperature is                 [2023]

  • 1×10-4

     

  • 1×10-6

     

  • 1×10-5

     

  • 1×10-3

     

(3)

Ka=Cα2

Ka=(0.1)×(0.01)2=1×10-5



Q 3 :    

The pKb of dimethylamine and pKa of acetic acid are 3.27 and 4.77 respectively at T(K). The correct option for the pH of dimethylammonium acetate solution is                [2021]

  • 6.25

     

  • 8.50

     

  • 5.50

     

  • 7.75

     

(4)

For a salt of weak acid and weak base

            pH=7+12(pKa-pKb)

Given, pKa=4.77,pKb=3.27

           pH=7+12(4.77-3.27)=7.75



Q 4 :    

Find out the solubility of Ni(OH)2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)2 is 2×10-15.            [2020]

  • 2×10-13 M

     

  • 2×10-8 M

     

  • 1×10-13 M

     

  • 1×108 M

     

(1)

Ni(OH)2sNi2+s+2OH-2s

where s is the solubility of Ni(OH)2.

NaOH0.1 MNa+0.1 M+OH-0.1 M

[OH-]=2s+0.10.1  ( 2s<<<0.1)

Ionic product of Ni(OH)2=[Ni2+][OH-]2=2×10-15=s(0.1)2

s=2×10-150.1×0.1=2×10-13 M



Q 5 :    

The pH of 0.01 M NaOH(aq) solution will be               [2019]

  • 7.01

     

  • 2

     

  • 12

     

  • 9

     

(3)

NaOH0.01 MNa++OH-0.01 M

  [OH-]=0.01 M

  pOH=-log[OH-]=-log(0.01)=2

  pH=14-pOH=14-2=12



Q 6 :    

Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:

A.  60 mL M10HCl + 40 mL M10 NaOH

B.  55 mL M10HCl + 45 mL M10 NaOH

C.  75 mL M5HCl + 25 mL M5 NaOH

D.  100 mL M10HCl + 100 mL M10 NaOH

pH of which one of them will be equal to 1?                                     [2018]

  • B

     

  • A

     

  • D

     

  • C

     

(4)

pH = 1, so [H+]=10-1

For acid-base mixture: N1V1-N2V2=N3V3

(For NaOH and HCl, Normality = Molarity)

A.   M1(H+)=60×110-40×110100=2×10-2 M  i.e. pH=1.6981.7

B.  M2(H+)=55×110-45×110100=1100=10-2 M  i.e. pH=2

C.  M3(H+)=75×15-25×15100=10-1M  i.e. pH=1

D.  M4(H+)=100×110-100×110200=0  i.e. pH=7



Q 7 :    

The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5NH) in a 0.10 M aqueous pyridine solution (Kb for C5H5N=1.7×10-9) is                [2016]

  • 0.0060%

     

  • 0.013%

     

  • 0.77%

     

  • 1.6% 

     

(2)

C5H5N0.10 M+H2OC5H5N+H+OH-

α=KbC=1.7×10-90.10=1.30×10-4

  Percentage of pyridine that forms pyridinium ion=1.30×10-4×100=0.013%



Q 8 :    

What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?          [2015]

  • 2.0

     

  • 7.0

     

  • 1.04

     

  • 12.65

     

(4)

One mole of NaOH is completely neutralised by one mole of HCl.

Hence, 0.01 mole of NaOH will be completely neutralised by 0.01 mole of HCl.

NaOH left unneutralised = 0.1 - 0.01 = 0.09 mol

As equal volumes of two solutions are mixed,

[OH]-=0.092=0.045 M

pOH=-log(0.045)=1.35    pH=14-1.35=12.65



Q 9 :    

Which of the following salts will give highest pH in water?              [2014]

  • KCl

     

  • NaCl

     

  • Na2CO3

     

  • CuSO4

     

(3)

Na2CO3, which is a salt of NaOH (strong base) and H2CO3 (weak acid), will produce a basic solution with pH greater than 7.