2-5 mL of 0-4 M weak monoacidic base pKb-12 at 25oC is titrated against 215 M HClaq- at 25oC- pKw-14 at 25oCThe hydrolysis constant is-

The correct option is B 0.01 Since #160; K _ h = K _ w K _ b = 10 ^ 14 K _ b #160; #160; #160; #160; #160; #160; #160; # ...

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Byju's Answer

Standard XII

Chemistry

Ostwald Dilution Law Mathematical Interpretation

2.5 mL of 0...

Question

$2.5 m L$ of $0.4 M$ weak monoacidic base ( $p K_{b} = 12$ at $25^{o} C$ ) is titrated against $\frac{2}{15} M$ $H C l$ (aq.) at $25^{o} C$ . ( $p K_{w} = 14$ at $25^{o} C$ )
The hydrolysis constant is:

1

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2

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5

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0.01

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Solution

The correct option is B $0.01$
Since $K_{h} = \frac{K_{w}}{K_{b}} = \frac{10^{- 14}}{K_{b}}$
$∵ (p K_{b} = - {l o g}_{10} K_{b} ∴ K_{b} = 10^{- 12})$
$↓$ $↓$
$★$ (It is constant $= \frac{10^{- 14}}{10^{- 12}} = 10^{- 2} = 0.01$
do not depend
on concentration)

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Similar questions

2.5 m L

2 / 5 M

weak monoacidic base (

K_{b} = 1 \times 10^{- 12})

25^{o} C

. The concentration of

H^{+}

ion at equivalence point is:
(

K_{w} = 1 \times 10^{- 14}

25^{o} C

)

2.5 m L

\frac{2}{5} M

weak monoacidic base

(K_{b} = 1 \times 10^{- 12}

25^{0} C)

is titrated with

\frac{2}{15} M H C l

in water at

25^{o} C

. The concentration of

H^{+}

at equivalence point is

(K_{w} = 1 \times 10^{- 14}

25^{0} C)

2.5 m L

\frac{2}{5} M

weak monoacidic base

(K_{b} = 1 \times 10^{- 12})

25^{o} C

is titrated with

\frac{2}{15} M

H C l

in water at

25^{o} C .

The concentration of

H^{+}

at equivalence points is

(K_{w} = 1 \times 10^{- 14} a t 25^{o} C)

2.5 m l

\frac{2}{5} M

weak monoacidic base

(K_{b} = {1 \times 10}^{- 12}

25^{0} C)

is titrated with

\frac{2}{5} M

HCl in water at

25^{0} C

. The concentration of

H^{+}

at equivalence point is: (

{K_{w} = 10}^{- 14}

25^{0} C

)

Q. 2.5 mL of

\frac{2}{5} M

weak monoacidic base

(K_{b} = 1 \times 10^{- 12} a t 25^{\circ} C)

is titrated with

\frac{2}{15} M H C l

in water at

25^{\circ} C

. The

p H

at equivalence point is

(K_{w} = 1 \times 10^{- 14} a t 25^{\circ} C)

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2.5 mL of 0.4 M weak monoacidic base (pKb=12 at 25oC) is titrated against 215 M HCl (aq.) at 25oC. (pKw=14 at 25oC)The hydrolysis constant is:

The correct option is B 0.01Since Kh=KwKb=10−14Kb ∵(pKb=−log10Kb∴Kb=10−12) ↓ ↓★ (It is constant =10−1410−12=10−2=0.01 do not depend on concentration)

$2.5 m L$ of $0.4 M$ weak monoacidic base ( $p K_{b} = 12$ at $25^{o} C$ ) is titrated against $\frac{2}{15} M$ $H C l$ (aq.) at $25^{o} C$ . ( $p K_{w} = 14$ at $25^{o} C$ )
The hydrolysis constant is:

The correct option is B $0.01$
Since $K_{h} = \frac{K_{w}}{K_{b}} = \frac{10^{- 14}}{K_{b}}$
$∵ (p K_{b} = - {l o g}_{10} K_{b} ∴ K_{b} = 10^{- 12})$
$↓$ $↓$
$★$ (It is constant $= \frac{10^{- 14}}{10^{- 12}} = 10^{- 2} = 0.01$
do not depend
on concentration)