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

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

A base type indicator B differs in colour from its conjugate acid ( BH ' ) . The acidic form is red in colour while the basic form is blue. The human eye can sense blue colour distinctly when the ratio of blue form concentration to red form concentration is a b or more. However, the human eye can distinctly sense the red colour when the ratio of red form concentration to blue form concentration. is c d or more. Determine the pH range of solution in which human eyes cannot observe distinct red or blue colours. Take B as K eq .

Ostwald developed a theory of acid-base indicators which offers a simple explanation for the colour change with the variation of pH . According to this theory, hydrogen ion indicators are either weak organic acid or base. The undissociated, molecules have one colour and the ion furnished by it, on dissociation has another the colour. Indicators are either weak acids or weak bases and when dissolved in water their dissociated form acquires a colour different from that of the undissociated form. Consider a weak acid indicator of the general formula H In , where in represents indicator. The equilibrium established in aqueous solution will be HIn ( aq ) → H + ( aq ) + In − ( aq ) Red Green Let K In be the equilibrium constant. K In = [ H + ] [ In − ] [ H In ] The human eye can detect the change in colour if the ratio of the two forms of indicator ranges between 0 . 1 to 10 . If, [ HIn ] [ In − ] = 1 . 0 the colour visible will be yellow. [ HIn ] [ In − ] = 10 , the colour visible will be red. [ HIn ] [ In − ] = 0 . 1 , the colour visible will be green. ∴ pOH = pK eq + log ( [ Acidic form ] [ Basic form ] ) = pK eq + log ( [ Red ] [ Blue ] ) ( pOH ) blue = pK eq + log ( b a ) p ( OH ) red = pK eq + log ( c d ) So range = pH blue to pH red = 14 - pK eq - log ( b a ) to 14 − pK eq − log ( c d ) B = K eq 14 - pB - log ( b a ) to 14 − pB − log ( c d ) Hence pH ranges between 14 - pB - log ( b a ) to 14 − pB − log ( c d )

In the reaction: Ni 2 + + 6 H 2 O → Ni H 2 O 6 2 + ,

Ni 2 + behaves like a Lewis acid.

H 2 O behaves like a Lewis base.

Above cannot be called an acid-base reaction.

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.

HARD

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IMPORTANT

Earn 100

Upon adding $50 mL$ of $0.2 M KOH$ solution to $50 mL$ of $0.5 M$ solution of ethylene diamine $(K_{b 1} = 8 \times 10^{- 5}; K_{b 2} = 2.7 \times 10^{- 8}),$ calculate $pH$ of final solution

Important Questions on Equilibrium

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IMPORTANT

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}

MEDIUM

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IMPORTANT

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

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Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 152

pH = 2,

half of the indicator, thymol blue (an acid type indicator) is in unionised form. Find the percentage of indicator in unionised form in the solution with

[H^{+}] = 4 \times 10^{- 3} M

HARD

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 153

A base type indicator

B

differs in colour from its conjugate acid

(BH')

. The acidic form is red in colour while the basic form is blue. The human eye can sense blue colour distinctly when the ratio of blue form concentration to red form concentration is

\frac{a}{b}

or more. However, the human eye can distinctly sense the red colour when the ratio of red form concentration to blue form concentration. is

\frac{c}{d}

or more. Determine the

pH

range of solution in which human eyes cannot observe distinct red or blue colours. Take

B

K_{eq} .

HARD

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 154

Determine the maximum number of moles of

{MgF}_{2}

that can dissolve in

1000 L

of a buffer solution of

pH = 4 .

Given:

K_{sp}

{MgF}_{2} = \frac{9}{14} \times 10^{- 8}

and

K_{a}

HF = 3.5 \times 10^{- 4} .

Take

(2.17)^{3} = 10

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 155

$(a)$ Determine the concentration of ${NH}_{3}$ solution whose $1 L$ can dissolve $0.1$ mole ${CuCO}_{3}$ .

Given $: K_{sp}$ of ${CuCO}_{3} = 1.4 \times 10^{- 10}$ and $K_{f}$ ${[Cu {({NH}_{3})}_{4}]}^{2 +} = 2 \times 10^{13} .$ Take $\sqrt[4]{\frac{1000}{2.8}} = 4.4 .$ Assume no other reaction to take place.

$(b)$ An aqueous solution of a metal bromide, $MBr {}_{2}(0.05 M)$ is saturated with $H_{2} S .$ Calculate the minimum $pH$ at which the metal sulphide will be precipitated. Concentration of $H_{2} S$ in a saturated solution $= 0.1 M$ $[K_{a 1}$ of $(H_{2} S) = 10^{- 7}, K_{a 2}$ of $(H_{2} S) = 1.2 \times 10^{- 13}, K_{sp}$ for $MS = 6 \times 10^{- 21}]$ .

EASY

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Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-2>Q 156

In the reaction:

{Ni}^{2 +} + 6 H_{2} O \to {[Ni {(H_{2} O)}_{6}]}^{2 +}

Upon adding 50 mL of 0.2 M KOH solution to 50 mL of 0.5 M solution of ethylene diamine Kb1=8×10-5; Kb2=2.7×10-8, calculate pH of final solution

Important Questions on Equilibrium

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Upon adding $50 mL$ of $0.2 M KOH$ solution to $50 mL$ of $0.5 M$ solution of ethylene diamine $(K_{b 1} = 8 \times 10^{- 5}; K_{b 2} = 2.7 \times 10^{- 8}),$ calculate $pH$ of final solution