Q: The ionization constant of HF,HCOOH and HCN at 298K are 6.8×10−4,1.8×10−4 and 4.8×10−9 respectively. Calculate the ioniation constants of the corresponding conjugate base.

We know for conjugate base:Kb=kakwFor F− and HF:Kb =6.8×10−41×10−14=1.47×10−11For HCOO− and HCOOH:Kb =1.8×10−41×10−14=5.56×10−11For CN− and HCN:Kb =4.8×10−91×10−14=2.08×10−6

Q: The first ionization constant of H2S is 9.1×10−8. Calculate the concentration of HS− ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2S is 1.2×10−13, calculate the concentration of S2− under both conditions.

(i) To calculate [HS−] in absence of HCl:Let, [HS−]=x M.H2S⇌H++HS−The initial concentrations of H2S,H+ and HS− are 0.1 M, 0 M and 0 M respectively.Their final concentrations are 0.1-x M, x M and x M respectively.Ka=[H2S][H+][HS−]9.1×10−8=0.1−xx×xIn the denominator, 0.1-x can be approximated to 0.1 as x is very small.9.1×10−8=0.1x×xx=9.54×10−5M=[HS−](ii) To calculate [HS−] in presence of HCl:Let [HS−]=y M.H2S⇌H++HS−The initial concentrations of H2S,H+ and HS− are 0.1 M, 0 M and 0 M respectively.Their final concentrations are 0.1-y M, y M and y M respectively.Also, HCl⇌H++Cl−For HCl [H+]=[Cl−]=0.1MKa=[H2S][HS−][H+]Ka=0.1−yy(0.1+y)In the denominator, 0.1-y can be approximated to 0.1 and in the numerator, 0.1+y can be approximated to 0.1 as y is very small.9.1×10−8=0.1y×0.1y=9.1×10−8M=[HS−](iii) To calculate [S2−] in absence of 0.1 M HCl:HS−⇌H++S2−[HS−]=9.54×10−5MLet [S2−]=X M. [H+]=9.54×10−5 MKa=[HS−][H+][S2−] 1.2×10−13=9.54×10−59.54×10−5×XX=1.2×10−13M=[S2−](iv) To calculate [S2−] in presence of 0.1 M HCl:Let [S2−]=X′ M.[HS−]=9.1×10−8 M[H+]=0.1 M (from HCl)Ka=[HS−][H+][S2−]1.2×10−13=9.1×10−80.1×X′X′=1.092×10−19=[S2−]

Q: What is the pH of 0.001 M aniline solution ? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Let c M be the initial concentration of [C6H5NH3+] and x be the degree of ionization.c(1−x)PhNH2+H2O⇌cxPhNH3++cxOH−Kb=[PhNH2][PhNH3+][OH−]=c(1−x)(cx)(cx)=1−xcx2But x is very small. Hence, 1−x≈1.Hence, Kb=cx2.The degree of ionization of aniline, x=cKb =0.0014.3×10−10=6.56×10−4[OH−]=cx=0.001×6.56×10−4=6.56×10−7M[H+]=[OH−]Kw=6.56×10−710−14=1.52×10−8pH=−log[H+]=−log1.52×10−8=7.818 Ka (conjugate base of aniline) =KbKw=4.3×10410−14=2.32×10−5

Q: The degree of dissociation of water at 25∘C is 1.9×10−7% and density is 1.0 g cm−3. The ionic constant for water is:

H2O ⇌ H+ + OH−(1−α)c αc αcα=1.9×10−7; Density of water =1.0 g/cm3∴c=181×1000=55.56 mol/L∴[H+]=55.56×1.9×10−9=1.055×10−7∴Kw=[H+][OH−]=(1.055×10−7)2=1.0×10−14

Q: What fraction of indicator is in acidic form at a solution pH of 5 , if pKa of the indicator is 4.7?

pH=pKln+log[Hln][ln−]⇒5=4.7+log[HIn][ln−]⇒0.3=log[Hln][ln−]⇒log2=log[HIn][ln−]⇒2=[HIn][ln−]Fraction =ln−+HIn HIn =31

Q: To a 200 mL of 0.1 M weak base BOH solution, 46 mL of 0.2 M solution of HCl are added. Now, what volume (in mL ) of 0.1 M H2SO4 should be added into above solution, so that pH of resulting solution becomes 9? [Kb(BOH)=10−5]

BOH+H+⟶B++H2Ot=0t=tt=50% neutralization (∵pOH=pKb=5)2010.8109.20009.21000∴0.8 millimole more of BOH are to be neutralized by H2SO4. ∴0.8=(0.1×2)×VmL∴VmL=4 mL

Question 1

The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

Accepted Answer

The pH of a solution is the negative logarithm of hydrogen ion concentration.pH=−log[H+] 3.76=−log[H+][H+]=1.7×10−4M

Question 2

The ionization constant of HF,HCOOH and HCN at 298K are 6.8×10−4,1.8×10−4 and 4.8×10−9 respectively. Calculate the ioniation constants of the corresponding conjugate base.

Accepted Answer

We know for conjugate base:Kb​=ka​kw​​For F− and HF:Kb​ =6.8×10−41×10−14​=1.47×10−11For HCOO− and HCOOH:Kb​ =1.8×10−41×10−14​=5.56×10−11For CN− and HCN:Kb​ =4.8×10−91×10−14​=2.08×10−6

Question 3

The first ionization constant of H2​S is 9.1×10−8. Calculate the concentration of HS− ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2​S is 1.2×10−13, calculate the concentration of S2− under both conditions.

Accepted Answer

(i) To calculate [HS−] in absence of HCl:Let, [HS−]=x M.H2​S⇌H++HS−The initial concentrations of H2​S,H+ and HS− are 0.1 M, 0 M and 0 M respectively.Their final concentrations are 0.1-x M, x M and x M respectively.Ka​=[H2​S][H+][HS−]​9.1×10−8=0.1−xx×x​In the denominator, 0.1-x can be approximated to 0.1 as x is very small.9.1×10−8=0.1x×x​x=9.54×10−5M=[HS−](ii) To calculate [HS−] in presence of HCl:Let [HS−]=y M.H2​S⇌H++HS−The initial concentrations of H2​S,H+ and HS− are 0.1 M, 0 M and 0 M respectively.Their final concentrations are 0.1-y M, y M and y M respectively.Also, HCl⇌H++Cl−For HCl [H+]=[Cl−]=0.1MKa​=[H2​S][HS−][H+]​Ka​=0.1−yy(0.1+y)​In the denominator, 0.1-y can be approximated to 0.1 and in the numerator, 0.1+y can be approximated to 0.1 as y is very small.9.1×10−8=0.1y×0.1​y=9.1×10−8M=[HS−](iii) To calculate [S2−] in absence of 0.1 M HCl:HS−⇌H++S2−[HS−]=9.54×10−5MLet [S2−]=X M. [H+]=9.54×10−5  MKa​=[HS−][H+][S2−]​ 1.2×10−13=9.54×10−59.54×10−5×X​X=1.2×10−13M=[S2−](iv) To calculate [S2−] in presence of 0.1 M HCl:Let [S2−]=X′ M.[HS−]=9.1×10−8  M[H+]=0.1 M (from HCl)Ka​=[HS−][H+][S2−]​1.2×10−13=9.1×10−80.1×X′​X′=1.092×10−19=[S2−]

Question 4

It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa​.

Accepted Answer

pH=−log[H+]=4.15[H+]=antilog(−4.15)=7.08×10−5[A−]=[H+]=7.08×10−5The concentration of undissociated acid is 0.01−0.000071=0.009929M.HA+H2​O⇌H3​O++A−Ka​=[HA][H3​O+][A−]​=0.009929(7.08×10−5)(7.08×10−5)​=5.05×10−7pKa​=−logKa​=−log5.05×10−7≈6.3

Question 5

The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa​ of bromoacetic acid.

Accepted Answer

[H+]=cα=0.1×0.132=0.0132MpH=−log[H+]=−log0.0132=1.88The acid dissociation constant is Ka​=1−αcα2​=1−0.1320.1×(0.132)2​=2.01×10−3.pKa​=−logKa​=−log(2.01×10−3)≈2.7

Question 6

What is the pH of 0.001 M aniline solution ? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Accepted Answer

Let c M be the initial concentration of [C6​H5​NH3+​] and x be the degree of ionization.c(1−x)PhNH2​​+H2​O⇌cxPhNH3+​​+cxOH−​Kb​=[PhNH2​][PhNH3+​][OH−]​=c(1−x)(cx)(cx)​=1−xcx2​But x is very small. Hence, 1−x≈1.Hence, Kb​=cx2.The degree of ionization of aniline, x=cKb​​​ =0.0014.3×10−10​​=6.56×10−4[OH−]=cx=0.001×6.56×10−4=6.56×10−7M[H+]=[OH−]Kw​​=6.56×10−710−14​=1.52×10−8pH=−log[H+]=−log1.52×10−8=7.818 Ka​ (conjugate base of aniline) =Kb​Kw​​=4.3×10410−14​=2.32×10−5

Question 7

The degree of dissociation of water at 25∘C is 1.9×10−7% and density is 1.0 g cm−3. The ionic constant for water is:

Accepted Answer

H2​O    ⇌     H+  +  OH−(1−α)c              αc            αcα=1.9×10−7; Density of water =1.0 g/cm3∴c=181​×1000=55.56 mol/L∴[H+]=55.56×1.9×10−9=1.055×10−7​∴Kw​=[H+][OH−]=(1.055×10−7)2=1.0×10−14

Question 8

What fraction of indicator is in acidic form at a solution pH of 5 , if pKa of the indicator is 4.7?

Accepted Answer

pH=pKln​+log[Hln][ln−]​⇒5=4.7+log[HIn][ln−]​⇒0.3=log[Hln][ln−]​⇒log2=log[HIn][ln−]​⇒2=[HIn][ln−]​Fraction =ln−+HIn HIn ​=31​

Question 9

To a 200 mL of 0.1 M  weak base BOH solution, 46 mL of 0.2 M solution of HCl are added. Now, what volume (in mL ) of 0.1 M H2​SO4​ should be added into above solution, so that pH of resulting solution becomes 9? [Kb​(BOH)=10−5]

Accepted Answer

BOH+H+⟶B++H2​Ot=0t=tt=50% neutralization (∵pOH=pKb​=5)​2010.810​9.200​​09.210​00​∴0.8 millimole more of BOH are to be neutralized by H2​SO4​. 
∴0.8=(0.1×2)×VmL​∴VmL​=4 mL

Question 10

The dissociation constant of a weak acid HA is 4.9×10−8. After making the necessary approximations, calculate  pH in a decimolar solution of the acid. Water has a pH of 7 .

Accepted Answer

From the dissociation of weak acid HA, HA⇌H++A−It α is the degree of ionization of the acid HA, then [H+]=0.1α[∵ the acid is decimolar ] [A−]=0.1α;[HA]=0.1(1−α)Therefore, 
Ka​=[HA][H+][A−]​=0.1(1−α)0.1α×0.1α​=0.10.1α×0.1α​( since acid is weak, 1−α=1)Ka​=0.1α2 or 4.9×10−8=0.1α2 or α2=0.14.9×10−8​ or α=7×10−4∴ Percentage ionization =100×7×10−4=7×10−2%

Calculation of pH: [H+]=0.1α=0.1×7×10−4 mole/ litre =7×10−5 mole/ litre Now since pH=​−log[H+]=−log[7×10−5]=5−log7=5−0.8451=4.1549​

Question 11

The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA,1M) is 1/100th  of that of a strong acid (HX,1M), at 25∘C. The Ka​ of HA is

Accepted Answer

Rate of hydrolysis in weak acid =1001​× Rate of hydrolysis in strong acid.[H+]weak acid =1001​×[H+]strong acid =1001​×1M=10−2M[H+]=Cα=0.01M1(α)=0.01α=0.01Ka​=Cα2=1(0.01)2=10−4

Question 12

The ionization constant of a weak electrolyte (HA) is 4×10−6 while the equivalent conductance of its 0.01M solution is 19.6 S cm2eq−1. The equivalent conductance of the electrolyte at infinite dilution (in S cm2 eq−1) will be ∧∘. Find the value of 10∧∘​.

Accepted Answer

Ka​=4×10−6          ∧eq​=19.6 S cm2eq−1C=0.01M                 Ka​=0.01×α2                 Ka​=0.01×α2Ka​=0.01×α2⇒α=10−24×10−6​​=2×10−2α=2×10−2=∧eq∘​19.6​Λeqo​=2×10−219.6​=980 S cm2 eq−1 10Λeqo​​=98

Question 13

A weak acid HA has Ka​ of 1.00×10−5. If 0.100 mol of this acid is dissolved in one litre of water, the pH of the solution is closed to

Accepted Answer

HA​⇌H++A− At eq.C(1−α)Cα+Cα Ka​=C(1−α)C2α2​(1>>>α) Ka​=Cα2 Ka​C=(Cα)2 H+=Cα=Ka​C​ H+=10−3 pH=3

Question 14

0.1 mol HCl is dissolved in distilled water of volume V then at lim v→∞ (pH)soluton​ is equal to

Accepted Answer

Kw​=[H+][OH−]
If we dilute 0.1MHCl to infinite volume than after a certain stage, [H+]<
10−7M
 Then [H+]=10−7M+10−xM where x>7 
So pH=7

Question 15

pH of solution of mixture of 0.1M 100 mL CH3​COOH(Ka​=2×10−5) with 0.1M,100 mL HCl will be :-

Accepted Answer

pH=−log[H+]
H+ from HCl and CH3​COOH
Acetic acid is weak .So, it is neglected
[H+]=[H+]HCl​.
conc=n/v=0.1= n/100= 10millimoles
total volume=100(HCl)+100(CH3​COOH)=200 ml
H+=n/v=10/200=0.05
pH=−log[0.05]
pH=1.30

Question 16

A weak acid react with strong base, ionization constant of weak acid is 10−4. Find out equilibrium constant for this reaction :-

Accepted Answer

HA⇌H++A−
 ionic product Ka​=[HA][H+][A−]​
strong base solution

HA+OH−⇌H2​O+A−. 
K=[OH−][HA][A−][H2​O]​
Kw​=[H+][OH−]=10−14
K=Kw​Ka​​=10−1410−4​=1010

Question 17

Sulphurous acid (H2​SO3​) has Ka1​=1.7×10−2 and Ka2​=6.4×10−8. The pH of 0.588MH2​SO3​ is.......... (Round off to the Nearest Integer).

Accepted Answer

For H2​SO3​Ka1​≫Ka2​ So we consider mainly 1st ionisation​t=00.588−H++HSO3Θ​t=teq.0.588(1−α)0.588α0.588αKa1​=1−α0.588α2​=1.7×10−2⇒α2=0.5881.7×10−2​α2=2.89×10−2α=2.89×10−2​α=1.7×10−1α=0.17[H+]=0.588×0.17=9.9×10−2pH=2−log9.9=1pH=1​

Question 18

An acid HA ionises as HA⇌H++A− The pH of 1.0M solution is 5 . Its dissociation constant would be

Accepted Answer

Exp. (a)
pH of a solution =5
∴[H+]=10−pH=10−5
HA (a weak acid) ionises as
HA⇌H++A−
1.000
(1−10−5)10−510−5
Ka​=[HA][H+][A−]​=(1−10−5)10−5×10−5​=110−10​
[∵1≫>>10−5]
=1×10−10M

Question 19

The dissociation constant of a weak acid is 4.9×10−8, its percentage ionisation at 0.1M is

Accepted Answer

x=cKa​​​=0.14.9×10−8​​=7×10−4 % ionisation =100x=7×10−2

Question 20

Ionisation constant of acetic acid is 1.8×10−5. The concentration of H+ions in 0.1M solution is

Accepted Answer

[H3​O+]=Ka​C​=1.8×10−5×0.1​=1.34×10−3M

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