The solution with p H value close to 1 is

75 m L o f 0 . 2 M H C l + 25 m L o f 0 . 2 M N a O H

10 m L o f 0 . 1 M H C l + 90 m L o f 0 . 1 M N a O H

55 m L o f 0 . 1 M H C l + 45 m L o f 0 . 1 M N a O H

75 m L o f 0 . 2 M H C l + 25 m L o f 0 . 1 M N a O H

Upon adding 50 mL of 0 . 2 M KOH solution to 50 mL of 0 . 5 M solution of ethylene diamine K b 1 = 8 × 10 - 5 ; K b 2 = 2 . 7 × 10 - 8 , calculate pH of final solution

Number of moles of KOH n KOH = 50 × 0 . 2 1000 = 0 . 01 mol Number of moles of C 2 H 8 N 2 n C 2 H 8 N 2 = 50 × 0 . 5 1000 = 0 . 025 mol Moles of salt = 50 × 0 . 25 1000 = 0 . 0125 mol Moles of left base = 0 . 025 − 0 . 0125 = 0 . 0125 mol pOH = pKb + log Salt Base pOH = − logK b 2 + log 0 . 0125 0 . 0125 pOH = - log ( 2 . 7 × 10 - 8 ) pOH = 7 . 5 PH = 14 - 7 . 5 pH = 6 . 5

The exothermic formation of ClF 3 is represented by the following equation: Cl 2 g + 3 F 2 g ⇌ 2 ClF 3 g ; ΔH = - 329 kJ Which of the following will increase the quantity of ClF 3 in the equilibrium mixture of Cl 2 , F 2 and ClF 3 ?

Increasing the volume of the container.

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IMPORTANT

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$C M$ fluoroacetic acid solution was found to contain $[H^{+}] = 1.5 \times 10^{- 3} M . K_{a}$ of fluoroacetic acid $= 2.5 \times 10^{- 3}$ . Then:

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Important Questions on Equilibrium

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 144

The solution with

p H

value close to

1

MEDIUM

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Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 145

Which of the following statement/s is/are correct about weak acids in aqueous solutions?

I. When $pH = {pK}_{a}$ of a monoprotic acid, $50 %$ of the acid is ionised
Il. If $pH = {pKa}_{2}$ of a diprotic acid, the average charge of all the ionised species is $0.5$
Ill. When $pH = pKa + 1, 10 %$ of the acid is ionised
IV. When $pH = 7, 50 %$ of a monibasic acid is ionized.

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Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 146

(a)

For which of the following, water would behave as a levelling solvent and for which, it would behave as a differentiating solvent? Strong acids, weak acids, strong bases, weak bases.

(b)

For which of the following, glacial acetic acid would behave as a levelling solvent and for which, it would behave as a differentiating solvent? Strong acids, weak acids, strong bases, weak bases.

(c)

For which of the following, liquid ammonia would behave as a levelling solvent and for which, it would behave as a differentiating solvent? Strong acids, weak acids, strong bases, weak bases.

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Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 147

Calculate

pOH, [{CH}_{3} {NH}_{3}^{+}], [C_{4} H_{13} N_{2}^{+}] & [C_{4} H_{14} N_{2}^{2 +}]

in an aqueous solution consisting of

0.2 M

methyl amine

(K_{b} = 4.1 \times 10^{- 4})

and

0.1 M

butane

- 1, 4

-

diaine

(K_{b 1} = 6.2 \times 10^{- 4} M; K_{b 2} = 2.25 \times 10^{- 5})

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Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 148

Upon adding

50 mL

0.2 M KOH

solution to

50 mL

0.5 M

solution of ethylene diamine

(K_{b 1} = 8 \times 10^{- 5}; K_{b 2} = 2.7 \times 10^{- 8}),

calculate

pH

of final solution

HARD

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Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 149

Determine the

pH

0.5 M {BH}_{2} {Cl}_{2}

solution (salt of a diacidic base

B

). Also calculate

[{BH}^{+}] & [B] .

Given

K_{b_{1}} & K_{b_{2}}

for

B

are

10^{- 6} & 2 \times 10^{- 10}

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Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 150

Prove that buffer capacity of

0.2 M {CH}_{3} COOH - 0.2 M {CH}_{3} COONa

buffer is less than the

0.4 M {CH}_{3} COOH - 0.4 {MCH}_{3} COONa

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Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 151

A small quantity of phenolphthalein (an acid indicator) is added to a decimolar solution of sodium butyrate. Calculate the ratio of the coloured to the colourless form of the indicator.

K_{a}

for butyric acid

1.5 \times 10^{- 5}, K

for the indicator

= 3.075 \times 10^{- 10}

and

K_{w} = 10^{- 14}

. Take

\log 1.23 = 0.09

C M fluoroacetic acid solution was found to contain H+=1.5×10-3 M. Ka of fluoroacetic acid =2.5×10-3. Then:

Important Questions on Equilibrium

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$C M$ fluoroacetic acid solution was found to contain $[H^{+}] = 1.5 \times 10^{- 3} M . K_{a}$ of fluoroacetic acid $= 2.5 \times 10^{- 3}$ . Then: