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Ncert Solutions Chemistry Class 11th

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6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

7.60. The pHof 0.1M of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

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Vishal Baghel

Contributor-Level 10

c=0.1M

pH= −log [H⁺]

=> 2.34 = −log [H⁺]

So, −log [H⁺]= 2.34

=> [H⁺]= 4.5*10⁻³

Also,

[H⁺]= cα

=>4.5*10⁻³= 0.1*α

=> α=4.5*10⁻²= 0.045

Then,

Ka = α²/c = (45*10⁻³)²/ 0.1

=202.5*10⁻⁶

=2.02*10⁻⁴

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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

7.59. The ionization constant of propanoic acid is 1.32 * 10^–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

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Vishal Baghel

Contributor-Level 10

Let the degree of ionization of propanoic acid be α. Then, representing propionic acid as HA, we have:

HA + H₂O ⇔ H₃O⁺ + A⁻

(.05−0.0α)≈0.5 0.05α 0.05α

K_a= [H₃O⁺] [A⁻] / [HA]

= (0.05α) (0.05α) / 0.05

= 0.05α²

=> α = (K_a /0.05)^1/2

^&nbs

...more

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6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

7.58. The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

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Vishal Baghel

Contributor-Level 10

Solubility of Sr (OH)₂=19.23g/L

The molecular weight of Sr (OH)₂ is 87.6 + 2 (17)=121.6

Then, concentration of Sr (OH)₂=19.23 /121.63M=0.1581M

Sr (OH)₂ (aq)→Sr²⁺ (aq)+2 (OH⁻) (aq)

∴ [Sr²⁺]=0.1581M

[OH⁻]=2*0.1581M=0.3126M

Now, K_w= [OH⁻] [H⁺]

=> [H+] = 10⁻¹⁴ / 0.3126

=> [H+]=3.2*10⁻¹⁴

∴pH= 13.495

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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

7.57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

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Vishal Baghel

Contributor-Level 10

[KOH]= [K⁺]= [OH⁻]= (0.561*1000) / (56*200)? =0.050M
[H+]=Kw / [OH⁻]? =10⁻¹⁴ / 0.05? =2.0*10⁻¹³
pH=−log [H⁺]=−log (2.0*10⁻¹³)

=12.7

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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

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.

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Vishal Baghel

Contributor-Level 10

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 = 6.31*10⁻³M
Egg white: [H⁺]=10^−7.8=1.58*10⁻⁸M

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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

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. &

...more

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Vishal Baghel

Contributor-Level 10

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

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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

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?

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Vishal Baghel

Contributor-Level 10

K_b= 5.4*10⁻⁴

c= 0.02M

Then, α= (K_b /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, K_b= [ (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 (100 / 0.02) = 0.54%

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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

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?

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Vishal Baghel

Contributor-Level 10

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 concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration

[H⁺] = 0.01 + x
The acetate ion concentration is equal to the hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.
[CH_3?

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6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

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.

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Vishal Baghel

Contributor-Level 10

Let c be the initial concentration of C₆H₅NH₃⁺ and x be the degree of ionisation.

C₆H₅NH₂+ H₂O? C₆H₅NH₃⁺+ OH^-

c (1-x) cx cx

K_b = [C₆H₅NH₃⁺] [ OH^-] / [C₆H₅NH₂]

= [cx] [cx] / [c (1 – x)]

Since x is very small and negligible 1 – x≈ 1

∴K_b= [cx] [cx] / [c] = cx²

=> x = $\sqrt[]{\frac{K b}{c}}$

= 6.56 x 10^-4

∴ [OH^-] = cx = 0.001 x 6.56 x 10^-4= 6.56 x 10^-7 M

[H⁺]= Kw / [OH^-] = 10^-14 / 6.56 x 10^-7= 1.52 x 10^-8

pH= –log [H⁺] = –log1.52 x 1

...more

Let c be the initial concentration of C₆H₅NH₃⁺ and x be the degree of ionisation.

C₆H₅NH₂+ H₂O? C₆H₅NH₃⁺+ OH^-

c (1-x) cx cx

K_b = [C₆H₅NH₃⁺] [ OH^-] / [C₆H₅NH₂]

= [cx] [cx] / [c (1 – x)]

Since x is very small and negligible 1 – x≈ 1

∴K_b= [cx] [cx] / [c] = cx²

=> x = $\sqrt[]{\frac{K b}{c}}$

= 6.56 x 10^-4

∴ [OH^-] = cx = 0.001 x 6.56 x 10^-4= 6.56 x 10^-7 M

[H⁺]= Kw / [OH^-] = 10^-14 / 6.56 x 10^-7= 1.52 x 10^-8

pH= –log [H⁺] = –log1.52 x 10^-8= 7.818

∴ Ionization constant of the conjugate acid of aniline,

K_a = K_w / K_b

= 10^-14 / (4.3 x 10^-10)

= 2.32 x 10^-5

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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Equilibrium

7.51. The pH of 0.005 M codeine (C₁₈H₂₁N0₃) solution is 9.95. Calculate the ionization constant and PK_b.

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Vishal Baghel

Contributor-Level 10

pH = 9.95,

pOH = 14 – pH = 14 − 9.95 = 4.05
[OH⁻] = 10^−pOH = 10^−4.05 = 8.913 * 10⁻⁵

Codeine + H₂? O? Codeine^H+ + OH⁻

The ionization constant, K_b? = [CodeineH⁺] [OH⁻] / [codeine]?

= [ (8.913*10⁻⁵)* (8.913*10⁻⁵)] / 5*10⁻³

= 1.588*10⁻⁶.

pK_b? = −log (1.588*10⁻⁶)

= 5.8

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