Ionisation constant of CH3COOH is 1.7×10−5 and concentration of H+ions is 3.4×10−4. Then, find out initial concentration of CH3COOH molecules.

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Ionisation constant of {C}{H}_{{{3}}}{C}{O}{O}{H} is {1.7}{10}

Question

Ionisation constant of $C H_{3} C O O H$ is $1.7 \times 10^{- 5}$ and concentration of $H^{+} i o n s$ is $3.4 \times 10^{- 4}$ . Then, find out initial concentration of $C H_{3} C O O H$ molecules.

3.4 \times 10^{- 4}

3.4 \times 10^{- 3}

6.8 \times 10^{- 4}

6.8 \times 10^{- 3}

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Lucas Discussed

Ionisation constant of

C H_{3} C O O H

1.7 \times 10^{- 5}

and concentration of

H^{+} i o n s

3.4 \times 10^{- 4}

. Then, find out initial concentration of

C H_{3} C O O H

molecules.

5 mins ago

Discuss this question LIVE

5 mins ago

C H_{3} C O O H \Leftrightarrow C H_{3} C O O^{-} + H^{+}

K_{a} = \frac{[ C H _{3} C O O ^{-} ] [ H ^{+} ]}{[ C H _{3} C O O H ]}

Given that,

[C H_{3} C O O^{-}] = [H^{+}] = 3.4 \times 10^{- 4} M

K_{a}

for

C H_{3} C O O H = 1.7 \times 10^{- 5}

C H_{3} C O O H

is weak acid, so in it

[C H_{3} C O O H]

is equal to initial concentration. Hence,

1.7 \times 10^{- 5} = \frac{( 3.4 \times 10 ^{- 4} ) ( 3.4 \times 10 ^{- 4} )}{[ C H _{3} C O O H ]}

[C H_{3} C O O H] = \frac{3.4 \times 10 ^{- 4} \times 3.4 \times 10 ^{- 4}}{1.7 \times 10 ^{- 5}}

= 6.8 \times 10^{- 3} M

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Question Text	Ionisation constant of $C H_{3} C O O H$ is $1.7 \times 10^{- 5}$ and concentration of $H^{+} i o n s$ is $3.4 \times 10^{- 4}$ . Then, find out initial concentration of $C H_{3} C O O H$ molecules.
Answer Type	Text solution:1
Upvotes	150