0-1 mole arsenic acid H3AsO4 is dissolved in a 1L buffer solution of pH - 8- Which of the following options is-are correct- For arsenic acid- Ka1 - 2-5 - 10-4 - Ka2 - 5 - 10-8 - Ka3 - 2 - 10-13- sign denotes that the higher concentration is 100 times greater than the lower one or more

Since H_3AsO_4 is a polybasic acid and given that[H^+]=10^ 8 it will ionise like nbsp; H_3AsO_4⇋ H_2AsO_4^ +H^+ nbsp; Ka_1=[H_2AsO_4^ ][H^+][H_3AsO_4] [H_3A ...

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Byju's Answer

Standard XIII

Chemistry

Buffer Solutions

0.1 mole arse...

Question

$0.1$ mole arsenic acid $(H_{3} A s O_{4})$ is dissolved in a 1L buffer solution of $p H = 8$ . Which of the following options is/are correct? For arsenic acid: $K a_{1} = 2.5 \times 10^{- 4}, K a_{2} = 5 \times 10^{- 8}, K a_{3} = 2 \times 10^{- 13}$ .
(‘<<’ sign denotes that the higher concentration is 100 times greater than the lower one or more)

[H_{3} A s O_{4}] << [H_{2} A s O_{4}^{-}]

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[H_{2} A s O_{4}^{-}] << [H A s O_{4}^{2 -}]

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[H A s O_{4}^{2 -}] << [H_{2} A s O_{4}^{-}]

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[A s O_{4}^{3 -}] << [H A s O_{4}^{2 -}]

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Solution

The correct option is D $[A s O_{4}^{3 -}] << [H A s O_{4}^{2 -}]$
Since $H_{3} A s O_{4}$ is a polybasic acid and given that $[H^{+}] = 10^{- 8}$
it will ionise like
$H_{3} A s O_{4} ⇋ H_{2} A s O_{4}^{-} + H^{+}$
$K a_{1} = \frac{[H_{2} A s O_{4}^{-}] [H^{+}]}{[H_{3} A s O_{4}]}$
$\frac{[H_{3} A s O_{4}]}{[H_{2} A s O_{4}^{-}]} = \frac{[H^{+}]}{K a_{1}} = \frac{10^{- 8}}{2.5 \times 10^{- 4}}$
$[H_{3} A s O_{4}] << [H_{2} A s O_{4}^{-}]$
similarly for second ionisation
$\frac{[H A s O_{4}^{2 -}]}{[H_{2} A s O_{4}^{-}]} = \frac{[H^{+}]}{K a_{2}} = \frac{10^{- 8}}{5 \times 10^{- 8}} = \frac{1}{5}$
for third ionisation
$\frac{[H A s O_{4}^{2 -}]}{[A s O_{4}^{3 -}]} = \frac{[H^{+}]}{K a_{3}} = \frac{10^{- 8}}{2 \times 10^{- 13}} = 5 \times 10^{4}$
Hence
$[A s O_{4}^{3 -}] << [H A s O_{4}^{2 -}]$

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Similar questions

0.1

mole arsenic acid

(H_{3} A s O_{4})

is dissolved in a 1L buffer solution of

p H = 8

. Which of the following options is/are correct? For arsenic acid:

K a_{1} = 2.5 \times 10^{- 4}, K a_{2} = 5 \times 10^{- 8}, K a_{3} = 2 \times 10^{- 13}

.
(‘<<’ sign denotes that the higher concentration is 100 times greater than the lower one or more)

Q. When

0.1 m o l

arsenic acid,

H_{2} A s O_{4}

is dissolved in

1 L

buffer solution of

p H = 4

, which of the following hold good?

K_{1} = 2.5 \times 10^{- 4}, K_{2} = 5 \times 10^{4}, K_{3} = 2 \times 10^{- 23}

for arsenic acid

[^{'} < <^{'}

sign denotes that the higher concentration is at least

100

times more than the lower one].

Q. To

100 m L

0.1 M

H_{3} P O_{4}

200 m L

0.1 M

N a O H

was added. Calculate the

p H

of the resulting solution.
(

G i v e n K_{a 1} = 10^{- 3}, K_{a 2} = 10^{- 8}, K_{a 3} = 10^{- 13}

)

25 m L

0.1 N H_{3} P O_{4}

is titrated against

0.1 N N a O H

. The

p H

of the solution after addition of

62.5 m L

of the base is :
(Given :

K_{a 1} = 1 \times 10^{- 3}, K_{a 2} = 1 \times 10^{- 8} a n d K_{a 3} = 1 \times 10^{- 13}

)

H_{3} A

is a weak triprotic acid

(K_{a_{1}} = 10^{- 5}, K_{a_{2}} = 10^{- 9}, K_{a_{3}} = 10^{- 13})

.
What is the value of

p X

0.1 M H_{3} A (a q .)

solution, where

p X = - log X

and

X = \frac{[A^{3 -}]}{[H A^{2 -}]}

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0.1 mole arsenic acid (H3AsO4) is dissolved in a 1L buffer solution of pH=8. Which of the following options is/are correct? For arsenic acid: Ka1=2.5×10−4,Ka2=5×10−8,Ka3=2×10−13. (‘<<’ sign denotes that the higher concentration is 100 times greater than the lower one or more)