Calculate the pH of 0-1 M NH4Cl solution in water -Kb NH4OH-1-8- 10-5- log1-8 - 0-25

Given, K_b=1.8×10^ 5 We know, pK_b= log(K_b) ∴ pK_b= log(1.8×10^ 5)=(5 0.25)=4.75 NH_4Cl is a salt of strong acid and weak base. pH=7 12(pK_b+logC) ⇒ pH=7 12(4. ...

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Standard XII

Chemistry

Introduction and Degree of Hydrolysis

Calculate the...

Question

Calculate the $p H$ of $0.1 M$ $N H_{4} C l$ solution in water :
$K_{b} (N H_{4} O H) = 1.8 \times 10^{- 5}, l o g 1.8 = 0.25$

4.5

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7.63

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6.53

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5.13

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Solution

The correct option is D $5.13$
Given,
$K_{b} = 1.8 \times 10^{- 5} We know, p K_{b} = - l o g (K_{b}) ∴ p K_{b} = - l o g (1.8 \times 10^{- 5}) = (5 - 0.25) = 4.75$
$N H_{4} C l$ is a salt of strong acid and weak base.
$p H = 7 - \frac{1}{2} (p K_{b} + l o g C) \Rightarrow p H = 7 - \frac{1}{2} (4.75 + l o g (0.1)) \Rightarrow p H = 7 - \frac{3.75}{2} \Rightarrow p H = 5.125$

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Similar questions

Q. Calculate the pH of that buffer soln which is obtained by dissolving 0.2 moles of

N H_{4} O H

& 0.25 mole of

N H_{4} C l

in water.

K_{b} = 1.8 \times 10^{- 5}

Q. Calculate the pH of the buffer solution containing

0.15 m o l e

{N H}_{4} O H

and

0.25 m o l e

{N H}_{4} C l

K_{b}

for

{N H}_{4} O H

1.8 \times 10^{- 5}

Q. Freshly precipitated

A l

and

M g

hydroxides are stirred vigorously in a buffer solution containing

0.25 M

{N H}_{4} C l

and

0.05 M

{N H}_{4} O H

[{A l}^{3 +}]

and

[{M g}^{2 +}]

in solution are:
Given:

K_{b}

for

{N H}_{4} O H = 1.8 \times 10^{- 5}

K_{{s p}_{A l {(O H)}_{3}}} = 6 \times 10^{- 32}

K_{{s p}_{M g {(O H)}_{2}}} = 8.9 \times 10^{- 12}

Q. Calculate amount of

N H_{4} C l

required to be dissolved in

500 m L

water to have

p H = 5. [K_{b} N H_{4} O H = 1.8 \times 10^{- 5}] .

(Molar mass of

N H_{4} C l = 53.5 g m o l^{- 1}

)

Q. Calculate the mass of

{N H}_{4} C l

dissolved in

500 m l

to have

p H = 4.5

. (

K_{b}

{N H}_{4} O H = 1.8 \times 10^{- 5})

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Calculate the pH of 0.1 M NH4Cl solution in water : Kb(NH4OH)=1.8×10−5,log1.8=0.25

The correct option is D 5.13Given, Kb=1.8×10−5We know, pKb=−log(Kb)∴ pKb=−log(1.8×10−5)=(5−0.25)=4.75 NH4Cl is a salt of strong acid and weak base. pH=7−12(pKb+logC)⇒pH=7−12(4.75+log(0.1))⇒pH=7−3.752⇒pH=5.125

Calculate the $p H$ of $0.1 M$ $N H_{4} C l$ solution in water :
$K_{b} (N H_{4} O H) = 1.8 \times 10^{- 5}, l o g 1.8 = 0.25$