Ncert Solutions Chemistry Class 11th
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7.56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
7.56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH=−log [H+]
Hence, [H+] = 10−pH
Milk: [H+] = 10−6.8 = 1.58×10−7M
Black coffee: [H+] = 10−5.0 =1×10−5M
Tomato juice: [H+] = 10−4.2 =6.31×10−5M
Lemon juice: [H+]=10−2.2 = 6.
Upvote 
7.55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2
(c) Human blood, 7.38 (d) Human saliva, 6.4.
7.55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2
(c) Human blood, 7.38 (d) Human saliva, 6.4.
a. Human muscle fluid 6.83
pH=6.83
pH=−log [H+]
∴6.83=−log [H+]
[H+]= 1.48 x 10−7M
b. Human stomach fluid, 1.2:
pH=1.2
1.2=−log [H+]
∴ [H+] = 0.063 M = 6.3 x 10-2 M
c. Human blood, 7.38:
pH=7.38=−log [H+]
∴ [H+]= 4.17 x 10−8M
d. Human saliva, 6.4:
pH=6.4
6.4=−log [H+]
[H+]= 3.98 x 10−7 M
Upvote 7.54. The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?
7.54. The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?
Kb= 5.4×10−4
c= 0.02M
Then, α= (Kb /c)1/2
α= (5.4×10−4 / 2 x 10-2)1/2 =0.1643
(CH3)2NH+H2O ↔ (CH3)2NH+2+OH-
[ (CH3)2NH] = 0.02 – x ≈ 0.02
[ (CH3)2NH+2] = x
[OH-] = 0.1 + x
≈ 0.1
Now, Kb= [ (CH3)2NH+2] [OH−]/ [ (CH3)2NH] = (x × 0.1) / (0.025).
x = 1.08 x 10-4
% of dimethylamine ionised = (1.08 x 10-4) x (10
Upvote 7.53. Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74.
How is the degree of dissociation affected when its solution also contains
(a) 0.01M (b) 0.1M in HCl?
7.53. Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74.
How is the degree of dissociation affected when its solution also contains
(a) 0.01M (b) 0.1M in HCl?
pKa? =? logKa= 4.74
Ka? = 10? pKa =10?4.74 = 1.8*10?5
Let x be the degree of dissociation. The concentration of acetic acid solution, C = 0.05 M
The degree of dissociation,
x= (Ka / C)1/2? = (1.8*10?5 / 0.05)1/2 ? = 0.019
(a) The solution is also 0.01 M in HCl.
Let x M be the hydrogen ion concentr
Upvote 7.52. 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.
7.52. 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 be the initial concentration of C6H5NH3+ and x be the degree of ionisation.
C6H5NH2 + H2O? C6H5NH3+ + OH-
c (1-x) cx
Upvote 7.51. The pH of 0.005 M codeine (C18H21N03) solution is 9.95. Calculate the ionization constant and PKb.
7.51. The pH of 0.005 M codeine (C18H21N03) solution is 9.95. Calculate the ionization constant and PKb.
pH = 9.95,
pOH = 14 – pH = 14 − 9.95 = 4.05
[OH−] = 10−pOH = 10−4.05 = 8.913 × 10−5
Codeine + H2? O? CodeineH+ + OH−
The ionization constant, Kb? = [CodeineH+] [OH−] / [codeine]?
= [ (8.913×10−5)× (8.913×10−5)] / 5×10−3
= 1.588×10−6.
pKb? = −log (1.588×10−6)
= 5.8
Upvote 7.50. 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.
7.50. 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.
H+] = cα = 0.1 × 0.132 = 0.0132M
pH = −log [H+] = −log0.0132
= 1.88
The acid dissociation constant is
Ka? = cα2? / (1−α) = 0.1 × (0.132)2 / (1−0.132)
= 2.01×10−3.
pKa? = −logKa? = −log (2.01×10−3) ≈ 2.7
Upvote 7.49. Calculate the pH of the following solutions:
(a) 2g ofTlOH dissolved in water to give 2 litre of the solution
(b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution
(c) 0.3 g of NaOH dissolved in water to give 200 mL of the solution
(d) l mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.
7.49. Calculate the pH of the following solutions:
(a) 2g ofTlOH dissolved in water to give 2 litre of the solution
(b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution
(c) 0.3 g of NaOH dissolved in water to give 200 mL of the solution
(d) l mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.
(a) For 2g of TlOH dissolved in water to give 2 L of solution:
[TlOH] = [OH−] = (2×1)? / (2×221) = (1 / 221)? M
pOH = −log [OH]− = −log (1/221)?
= 2.35
pH = 14 – pOH = 14 − 2.35 = 11.65
(b) For 0.3 g of Ca (OH)2? dissolved in water to give 500 mL of solution:
[OH−] = 2 [Ca (
Upvote 7.48. Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH.
7.48. Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH.
(a) 0.003 M HCl
[H3? O+] = [HCl] = 0.003M
pH = −log [H+] = −log (3.0×10−3) = 2.523
(b) 0.005 M NaOH
[OH−] = [NaOH] = 0.005M
[H+] = Kw? / [OH−]? = 10−14/ 0.005? =2×10−12
pH= −log [H+]=−log (2×10−12)=11.699
(c) 0.002M HBr
[H+]= [HBr]=0.002
pH= −log [H+]=−log0.002=2.699
(d) 0.002M KOH
[OH−]= [KOH]
Upvote 7.47. 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.
7.47. 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.
pH= −log [H+]=4.15
[H+]= antilog (−4.15)= 7.08×10−5
[A−]= [H+]=7.08×10−5
The concentration of undissociated acid is 0.01−0.000071=0.009929M.
HA+H2? O? H3? O++A−
Ka? = [H3? O+] [A−] / [HA]? = (7.08×10−5) (7.08×10−5)? / 0.009929
= 5.05×10−7
pKa? = −logKa? = −log5.05×10−7 ≈ 6.3
Upvote