$C_{2} H_{5} O H$ ionises as:

C_{2} H_{5} O H ⇌ C_{2} H_{5} O^{-} + H^{+}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 C_{2} H_{5} O H ⇌ C_{2} H_{5} O H_{2}^{+} + C_{2} H_{5} O^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{2} H_{5} O H ⇌ C_{2} H_{5}^{+} + O H^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{2} H_{5} O H + H_{2} O ⇌ C_{2} H_{5} O^{-} + H_{3} O^{+}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $C_{2} H_{5} O H ⇌ C_{2} H_{5} O^{-} + H^{+}$
$C_{2} H_{5} O H$ is slightly acidic because it donates its hydroxyl proton* to strong bases.
Therefore it ionises as
$C_{2} H_{5} O H ⇌ C_{2} H_{5} O^{-} + H^{+}$
*hydroxyl proton is the $H^{+}$ ion of an OH group.

Suggest Corrections

Similar questions

C_{2} H_{5} O H, {(C H}_{3})_{2} N H, C_{2} H_{5} {N H}_{2}

C_{2} H_{5} O H

{C H}_{3} O H

N a / C_{2} H_{5} O H

C H_{3} C O O C_{2} H_{5} + H_{2} O H^{+} - ----- \to C H_{3} C O O H + C_{2} H_{5} O H

C_{2} H_{4} + H_{2} O = C_{2} H_{5} O H

{Δ H}_{f o r m}^{o} (C_{2} H_{4}) = 55, 28 K J / m o l

{Δ H}_{f o r m}^{o} (H_{2} O) = - 241.8 K J / m o l

{Δ H}_{f o r m}^{o} (C_{2} H_{5} O H) = - 235.3 K J / m o l

Join BYJU'S Learning Program