Formic acid is 4-5- dissociated in a 0-1 N solution at 20- C- The ionization constant of formic acid is-

When formic acid is dissolved in water, it partially dissociates (4.5%). K_a=Cα× CαC(1 α ) We assume α∼ 0⇒ 1 α∼ 1 ⇒K_a=Cα ^21=0.1× 0.045× 0.045=2.02× 10^ 4 ...

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Standard XII

Chemistry

Arrhenius Bases

Formic acid i...

Question

Formic acid is $4.5 %$ dissociated in a $0.1 N$ solution at $20^{\circ} C$ . The ionization constant of formic acid is:

21 \times 10^{- 4}

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21

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0.21 \times 10^{- 4}

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2.1 \times 10^{- 4}

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Solution

The correct option is D $2.1 \times 10^{- 4}$
When formic acid is dissolved in water, it partially dissociates $(4.5 %)$ .
$K_{a} = \frac{C α \times C α}{C (1 - α)}$
We assume $α \sim 0 \Rightarrow 1 - α \sim 1$
$\Rightarrow K_{a} = \frac{C α^{2}}{1} = 0.1 \times 0.045 \times 0.045 = 2.02 \times 10^{- 4}$

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Formic acid is 4.5% dissociated in a 0.1 N solution at 20∘ C. The ionization constant of formic acid is:

The correct option is D 2.1×10−4When formic acid is dissolved in water, it partially dissociates (4.5%). Ka=Cα×CαC(1−α) We assume α∼0⇒1−α∼1 ⇒Ka=Cα21=0.1×0.045×0.045=2.02×10−4

Formic acid is $4.5 %$ dissociated in a $0.1 N$ solution at $20^{\circ} C$ . The ionization constant of formic acid is: