A mixture of weak acid is 01M in HCOOH K a -1-8- 10 -4 and 0-1M in HOCN K a -3-3- 10 -4 - Hence- - H 3 O - - is-

The correct option is A 7.14× 10 ^ 3 M For a weak acid, [H^+]=Cα=C·√(K_aC)=√(K_a· C) Hence, total [H^+]=(Cα)_HCOOH+(Cα)_HOCN =√(10^ 2× 1.8 ...

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Byju's Answer

Standard XII

Chemistry

Bronsted Lowry Theory

A mixture of ...

Question

A mixture of weak acid is 01M in $H C O O H$ $(K_{a} = 1.8 \times 10^{- 4})$ and 0.1M in $H O C N$ $(K_{a} = 3.3 \times 10^{- 4})$ . Hence, $[H_{3} O^{+}]$ is:

7.14 \times 10^{- 3} M

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4.1 \times 10^{- 4} M

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0.20 M

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4.1 \times 10^{- 3} M

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Solution

The correct option is A $7.14 \times 10^{- 3} M$
For a weak acid, $[H^{+}] = C α = C \cdot \sqrt{\frac{K_{a}}{C}} = \sqrt{K_{a} \cdot C}$
Hence, total $[H^{+}] = (C α)_{H C O O H} + (C α)_{H O C N}$
$= \sqrt{10^{- 2} \times 1.8 \times 10^{- 4}} + \sqrt{10^{- 1} \times 3.3 \times 10^{- 4}}$
$= (1.4 + 5.74) \times 10^{- 3} M$
$\approx 7.14 \times 10^{- 3} M$

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Similar questions

[H^{+}]

in a solution containing

0.1 M

H C O O H

and

0.1 M

H O C N

X \times 10^{- 3} M

. Nearest integer to X is

K_{a}

for

H C O O H

and

H O C N

are

1.8 \times 10^{- 4}

and

3.3 \times 10^{- 4}

respectively

Q. Calculate

[H^{+}]

in a solution containing

0.1

H C O O H

and

0.1

H O C N

K_{a}

for

H C O O H

and

H O C N

are

1.8 \times 10^{- 4}

and

3.3 \times 10^{- 4}

respectively.

Q. Calculate the pH of

0.1 M

solution of

{N H}_{4} O C N . K_{b}

for

{N H}_{3}

1.75 \times 10^{- 5}

and

K_{a}

for

H O C N

3.3 \times 10^{- 4}

Q. At

25, [H_{3} O^{+}] = 1 \times 10^{- 7} M

in water, hence,

K_{a}

is:

Q. The

[H^{+}]

of the resulting solution that is 0.01 M in acetic acid

(K_{a} = 1.8 \times 10^{- 5} M)

and 0.01 M in benzoic acid.

(K_{a} = 6.3 \times 10^{- 5} M)

will be:

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A mixture of weak acid is 01M in HCOOH (Ka=1.8×10−4) and 0.1M in HOCN (Ka=3.3×10−4). Hence, [H3O+] is:

The correct option is A 7.14×10−3MFor a weak acid, [H+]=Cα=C⋅√KaC=√Ka⋅CHence, total [H+]=(Cα)HCOOH+(Cα)HOCN=√10−2×1.8×10−4+√10−1×3.3×10−4=(1.4+5.74)×10−3M≈7.14×10−3M

A mixture of weak acid is 01M in $H C O O H$ $(K_{a} = 1.8 \times 10^{- 4})$ and 0.1M in $H O C N$ $(K_{a} = 3.3 \times 10^{- 4})$ . Hence, $[H_{3} O^{+}]$ is: