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8 months ago

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

8 months ago

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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8 months ago

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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8 months ago

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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8 months ago

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

8 months ago

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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8 months ago

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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8 months ago

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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New answer posted

8 months ago

Class 11th Equilibrium

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 PK_a of bromoacetic acid.

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Vishal Baghel

Contributor-Level 10

H⁺] = cα = 0.1 * 0.132 = 0.0132M

pH = −log [H⁺] = −log0.0132

= 1.88

The acid dissociation constant is

K_a? = cα²? / (1−α) = 0.1 * (0.132)² / (1−0.132)

= 2.01*10⁻³.

pK_a? = −logK_a? = −log (2.01*10⁻³) ≈ 2.7

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New answer posted

8 months ago

Class 11th Equilibrium

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)₂dissolved 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.

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Vishal Baghel

Contributor-Level 10

(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 (OH)₂? ] = 2 (0.3*1000/500? ) = 1.2M
pOH = −log [OH⁻] = −log1.2 = 1.79

pH= 14−pOH=14−1.79

=12.21

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

[OH⁻]= [NaOH] = 0.3*1000/200? = 1.5M
pOH= −log [OH⁻] = −log1.5 = 1.43

pH= 14 – pOH = 14 − 1.43

= 12.57

(d) For 1mL of 13.6 M HCl diluted w

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