You visited us 2 times! Enjoying our articles? Unlock Full Access!

Arrhenius Acid

In 0.1 M solu...

Question

In 0.1 M solution, weak electrolyte, $C H_{3} C O O H$ is 1.3% dissociated. If 100 molecules of $C H_{3} C O O H$ are placed originally in the solution, how many acid molecules remain nonionized at equilibrium?

Open in App

Solution

$C H_{3} C O O H \to C H_{3} C O O^{-} + H^{+}$
No. of acid molecules non ionized $= 100 - \frac{100 \times 1.3}{100}$
$= 98.7 ≃ 99$

Suggest Corrections

Similar questions

2.0 g

100 mL

0.05 M C H_{3} C O O H

C H_{3} C O O H

0.49 M

3 \times 10^{2} m^{2} g^{- 1}

{C H}_{3} C O O H

1.8 \times 10^{- 5}

{C H}_{3} C O O H

50 m l 0.1 M N a O H

50 m l

0.1 M

C H_{3} C O O H

p H

{C H}_{3} C O O H = 4.7447

log 20 = 1.3

^{2}

^{- 1}

{CH}_{3} COOH

C H_{3} C O O H

C H_{3} C O O H

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

C H_{3} C O O H ?

Join BYJU'S Learning Program