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Equilibrium

Short Answer Type

291.

Calculate the degree of hydrolysis of 0.015 M solution of NH₄Cl. Given K_b for NH₄OH is 1·8 × 10^–5, K_w = 10^–14at 25°C.

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Long Answer Type

292.

Explain the terms: buffer solution and buffer capacity.

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293.
Show with an example how buffer solution resists the action of acid or base towards change in pH.
Or
Discuss the buffer action of:
(i) acidic buffer
(ii) basic buffer.

The action of acidic buffer: Consider an equimolar mixture of CH₃COONa (strong electrolyte) and CH₃COOH (weak electrolyte).

Ionisation of acetic acid is suppressed by acetate ion[common ion effect].
There will be a large concentration of Na⁺ ions. CH₃COO^– ions and undissociated CH₃COOH molecules.
(i) When a few drops of acid (say HCl) are added to it, the H⁺ ion from the added acid combines with an excess of CH₃COO^– ion to form feebly ionised CH₃COO^– ion to form feebly ionised CH₃COOH.

Thus there is no increase in the concentration of H⁺ ions, pH remains constant.

(ii) When a few drops of a base (say NaOH) are added, OH^– ions of the added base are neutralised by the H⁺ ions (of buffer) to form feebly ionised water molecules.

Thus there is no increase in the concentration of OH^– ions and hence pH remains constant.
The reverse acidity is due to CH₃COOH and reserves basicity to due to CH₃COO^– ions.
The action of basic buffer: A basic buffer is a mixture (equimolar) of a weak base and its salt with a strong acid. Consider an equimolar mixture of NH₄OH (weak electrolyte) and NH₄Cl (strong electrolyte).

Ionisation of NH₄OH is suppressed by NH₄⁺ ions [common ion effect].
There will be a large concentration of NH₄⁺ ions, Cl^– ions and NH₄OH molecules.

(i) When a few drops of acid (say HCl) are added, the H⁺ ions of the added acid combine with OH^- ions of-NH₄OH (of buffer) to form feebly ionised water. Thus no change in pH occurs.

According to Le Chatelier’s principle, the above reaction results in the greater dissociation of NH₄OH to restore the original concentration of OH^– ions.

(ii) When a few drops of a base (say NaOH) are added to the buffer, the OH^– ions of added base combine with NH₄⁺ ions to form unionised NH₄OH.

As the concentration of OH^– ions does not increase, the pH value remains unchanged.
In this buffer, reserve acidity is due to NH₄⁺ ions and reserve alkalinity is due to NH₄OH molecules.

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294. How will you explain buffer action of aqueous solution of ammonium acetate?

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295.

Calculate the pH of:
(i) an acidic buffer mixture
(ii) a basic buffer mixture.
Or
Derive Henderson’s equation for an acidic and basic buffer mixture.
Or
Derive the following equation for the pH of an acidic buffer:
$pH space equals space pK subscript straight a space plus space log space fraction numerator open vertical bar Salt close vertical bar over denominator open vertical bar Acid close vertical bar end fraction$

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Short Answer Type

296. Calculate the pH value of a solution obtained by mixing 0·083 moles of acetic acid and 0·091 moles of sodium acetate and making the volume 500 ml. K_a for acetic acid is 1·75 × 10^–5.

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Long Answer Type

297. Calculate the pH of a solution obtained by mixing 6·0g of acetic acid and 12·30g of sodium acetate and making the volume of solution to 500 ml. K_a for acetic acid is 1·8 × 10^–5

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298. Determine the pH of a solution obtained by mixing equal volumes of 0·015N NH₄OH and 0·15N NH₄NO₃ solutions. (K_b for NH₄OH is 1·8 × 10^–5).

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299.

Describe Ostwald’s theory of acid-base indicators.
Or
How does Ostwald’s theory explain the colour change of:
(i) Phenolphthalein
(ii) Methyl orange in acid-base titrations?

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