The hydride ion $H^{-}$ is stronger base than hydroxide ion $O H^{-}$ . Which of the following reaction will occur if sodium hydride $(N a H)$ is dissolved in water?

H^{-} (a q .) + H_{2} O \to H_{3} O^{-}

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H^{-} (a q .) + H_{2} O (l) \to O H^{-} + H_{2}

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H^{-} + H_{2} O \to

No reaction

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None of the above

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Solution

The correct option is B $H^{-} (a q .) + H_{2} O (l) \to O H^{-} + H_{2}$
As hydride ion is stronger base, so it will react with water which acts as an acid. The reaction occurring is as follows:
$H^{-} (a q .) + H_{2} O (l) \to O H^{-} + H_{2}$ .

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Similar questions

In $H_{2} - O_{2}$ fuel cell the reaction occuring at cathode is:

Q. The conjugate base of

H_{2} S O_{4}

in the following reaction is:

H_{2} S O_{4} + H_{2} O \to H_{3} O^{+} + H S O_{4}^{-}

C H_{3} C H_{2} C l + K O H (a q .) \to C H_{3} C H_{2} - O H + K C l

The above reaction is:

C l_{2} (g) + O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q)

Balance the above reaction using oxidation number method.

From the following data, mark the option(s) where $Δ$ H is correctly written for the given reaction.

Given : $H^{+} (a q) + O H^{-} (a q) \to H_{2} O (ℓ); Δ H = - 57.3 k J$

$Δ H_{solution} of HA(g) = - 70.7 kJ/mol$

$Δ H_{solution} of BOH(g) = + 20 kJ/mol$

$Δ H_{ionization} of H A = 15 kJ/mol$ and BOH is a strong base.

$\begin{matrix} Reaction & Δ H_{r} (kJ/mol) (a) & HA(aq) + BOH(aq) \to BA(aq) + H_{2} O & - 42.3 (b) & HA(g) + BOH(g) \to BA(aq) + H_{2} O & - 93 (c) & HA(g) \to H^{+} (a q) + A^{-} (a q) & - 55.7 (d) & B^{+} (aq) + O H^{-} (aq) \to BOH(aq) & - 20 \end{matrix}$

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The hydride ion H− is stronger base than hydroxide ion OH−. Which of the following reaction will occur if sodium hydride (NaH) is dissolved in water?

The correct option is B H−(aq.)+H2O(l)→OH−+H2As hydride ion is stronger base, so it will react with water which acts as an acid. The reaction occurring is as follows: H−(aq.)+H2O(l)→OH−+H2.

The hydride ion $H^{-}$ is stronger base than hydroxide ion $O H^{-}$ . Which of the following reaction will occur if sodium hydride $(N a H)$ is dissolved in water?

The correct option is B $H^{-} (a q .) + H_{2} O (l) \to O H^{-} + H_{2}$
As hydride ion is stronger base, so it will react with water which acts as an acid. The reaction occurring is as follows:
$H^{-} (a q .) + H_{2} O (l) \to O H^{-} + H_{2}$ .