Q 1 :

Ka for CH3COOH is 1.8×10-5  and Kb for NH4OH is 1.8×10-5.  The pH of ammonium acetate solution will be _______ .    [2024]    



(7) 

As  Ka  of  CH3COOH=Kb  of  NH4OH

So, pKa  of  CH3COOH=pKb  of  NH4OH

And  pKa-pKb=0

For salt of weak acid and weak base:

pH=12(pKw+pKa-pKb)

pH=12(14+0)=7

 



Q 2 :

pH of water is 7 at 25°C. If water is heated to 80°C, its pH will :                [2025]

  • Remains the same

     

  • Decrease

     

  • Increase

     

  • H+ concentration increases, OH- concentration decreases

     

(2)

H2O⇌H++OH-

With increase in temperature, dissociation of water increases, hence [H+] increases. Increase of [H+] means decrease of pH (pH=-log[H+]).

 



Q 3 :

An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would

(Given log 2 = 0.30)                                         [2025]

  • reduce to 0.5

     

  • increase to 1.3

     

  • remain same

     

  • increase to 2

     

(2)

Initial concentration of H+([H+]initial)=10-pHinitial=10-1 M 

As equal volume is added, concentration of H+ becomes half

([H+]initial)=10-12 M

pHfinal=-log[H+]final =-log(10-12)

             =-log(10-1)+log2 =1+0.3=1.3



Q 4 :

A weak acid HA has degree of dissociation x. Which option gives the correct expression of (pH-pKa)?            [2025]
 

  • log(1+2x)

     

  • 0

     

  • log(x1-x)

     

  • log(1-xx)

     

(3)

HA⇌H++A-(t=0) (moles)a000(t=∞) (moles)a(1-x)a0x a0x 

Ka=[H+][A-][HA]=[H+]×a0xa0(1-x)

Ka=[H+]×x(1-x)

Taking -log on both sides

-logKa=-log([H+]×x(1-x))

-logKa=-log[H+]-log(x1-x)

pKa=pH-log(x1-x)

pH-pKa=log(x1-x)



Q 5 :

If 1 mM solution of ethylamine produces pH = 9, then the ionization constant (Kb) of ethylamine is 10-x. The value of x is ______ (nearest integer).

[The degree of ionization of ethylamine can be neglected with respect to unity.]                      [2025]



(7)

C2H5NH2(aq)+H2O⇌C2H5NH3+(aq)+OH-(aq)t=010-3 Mexcess00t=∞10-3(1-α)10-3αM10-3αM≃10-3 M

pOH=14-pH=14-9=5 

Thus [OH]-=10-5 M

Also, [C2H5NH3+]=[OH]-=10-5 M

Kb=[OH]-[C2H5NH3+][C2H5NH2]=10-5×10-510-3=10-7



Q 6 :

The pH of a 0.01 M weak acid HX (Ka=4×10-10) is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6.  The new concentration of the diluted weak acid is given as x×10-4 M.  The value of x is ______ (nearest integer).             [2025]



(25)

New concentration of H+([H+]new)=10-pH(new) M=10-6 M 

KaCnew=[H+]new

4×10-10 Cnew=10-6

4×10-10 Cnew=10-12 

Cnew=25×10-4 M

Note: In the question it should be 0.25 M weak acid instead of 0.01 M weak acid. However, this data is not needed to solve the question.



Q 7 :

x mg of Mg(OH)2 (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is ____ mg. (Nearest integer)  

(Given: Mg(OH)2 is assumed to dissociate completely in H2O)                    [2025]
 



(3)

pOH=14-pH=14-10=4 

[OH-]=10-pOH=10-4M 

Moles of hydroxide (nOH-)=[OH-]×volume 

=10-4M×1 L=10-4 mol 

Moles of Mg(OH)2(nMg(OH)2)=nOH-2=10-42mol

Mass of Mg(OH)2=nMg(OH)2×MMg(OH)2 

=10-42×58=29×10-4 g=2.9 mg≈3 g



Q 8 :

When the hydrogen ion concentration [H+] changes by a factor of 1000, the value of pH of the solution ______               [2023]

  • decreases by 3 units

     

  • decreases by 2 units

     

  • increases by 2 units

     

  • increases by 1000 units

     

(1)

(pH)1=-logC

(pH)2=-log103C=-[log103+logC]

           =-3-logC

(pH)2-(pH)1=-3 ⇒ Decreased by 3



Q 9 :

Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solutions of pH 12 is x×10-1. The value of x is ______ (Nearest integer). Assume complete dissociation.                   [2023]



(5)

pH=12⇒pOH=2  so  [OH-]=10-2 M.

Millimole of [OH-]=10-2×100=1

Ca(OH)2(s)⟶Ca2+(aq.)+2OH-(aq.)

12millimole                        1 millimole

millimole of Ca(OH)2=12=5×10-1millimole.



Q 10 :

600 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M H2SO4. The pH of the mixture is _____ ×10-2. (Nearest Integer)

[Given log 2 = 0.30

           log 3 = 0.48

           log 5 = 0.69

           log 7 = 0.84

           log 11 = 1.04]                                   [2023]



(186)

HCl⟶H++Cl-Milli moles0.01×6006H2SO4⟶2H++SO4-2Milli moles0.01×4002×4=8

[H+]=8+61000=14×10-3

      pH=-log(14×10-3)

             =3-log 14

             =3-log (2×7)

              =3-log 2-log 7

              =3-0.30-0.84

              =1.86=186×10-2



Q 11 :

Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with K4[Fe(CN)6] is:

Cu2+,Fe3+,Ba2+,Ca2+,NH4+,Mg2+,Zn2+                     [2025]



(3)

 



Q 12 :

4 gms of NaOH and 4.9 gm H2SO4 are dissolved in water and volume is made up to 250 ml. The pH of the solution is



(7)

No.of equation acid = no.of equation base



Q 13 :

The conjugate base for bicarbonate ion (HCO3-) is

  • CO32-

     

  • HCO3-

     

  • CO2

     

  • H2CO3

     

(1)

Conceptual



Q 14 :

The first and second ionization constants of a weak dibasic acid H2A are 8.1×10-8 and 1.0×10-13 respectively. 0.1 mol of H2A was dissolved in 1 L of 0.1 M HCl solution. The concentration of HA- in the resultant solution is:             [2026]

  • 0.1 M

     

  • 9.53×10-6 M

     

  • 8.1×10-8 M

     

  • 1.0×10-13 M

     

(3)

HCl→Cl-+H+t=0 (conc.)0.100t=∞ (conc.)00.10.1

H2A⇌HA-+H+t=0 (conc.)0.100t=∞ (conc.)0.1-x≃0.1xx

Due to common ion effect dissociation of H2A is suppressed, thus, [H+]total=x+0.1≃0.1,[H2A]=0.1-x≃0.1

8.1×10-8=x×0.10.1

x=8.1×10-8

Thus concentration of HA-=8.1×10-8 M



Q 15 :

Amongst the following the total number of compounds soluble in aqueous NaOH at room temperature is:            [2026]

[IMAGE 134]

  • 5

     

  • 4

     

  • 6

     

  • 3

     

(1)

Acidic strength order is : carboxylic acids > phenols > water > alcohols > amines

Compounds more acidic than water are soluble in aqueous NaOH as acid-base reactions proceed from strong acid/base to weak acid/base.

                 HAStronger acid+NaOH→NaA+H2OWeaker acid

Thus II, III, IV, V and VII are soluble in aqueous NaOH.



Q 16 :

Arrange the following resultant mixtures in increasing order of their pH values.

A. 10 mL 0.2 M Ca(OH)2+25 mL 0.1 M HCl

B. 10 mL 0.01 M H2SO4+10 mL 0.01 M Ca(OH)2

C. 10 mL 0.1 M H2SO4+10 mL 0.1 M KOH

Choose the correct answer from the options given below:                   [2026]

  • B<C<A

     

  • C<A<B

     

  • C<B<A

     

  • A<C<B

     

(3)

Solution mmol of OH- mmol of H+ Nature of resultant solution
A 0.2×10×2=4 0.1×25×1=2.5 Basic as OH- is in excess
B 0.01×10×2=0.2 0.01×10×2=0.2 Neutral
C 0.1×10×1=1 0.1×10×2=2 Acidic as H+ is in excess

 

Increasing order of pH=C<B<A