SATHEE: Equilibrium Question 50

Equilibrium Question 50

Question: $1 \times 10^{- 12}$ of $1 M H_{2} S O_{4}$ will completely neutralise

Options:

A) $10 m l$ of $1 M N a O H$ solution

B) $10 m l$ of $2 M N a O H$ solution

C) $5 m l$ of $2 M K O H$ solution

D) $5 m l$ of $1 M N a_{2} C O_{3}$ solution

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Answer:

Correct Answer: B

Solution:

$H_{2} S O_{4} + 2 H_{2} O$ ⇌ $2 H_{3} O^{+} + S O_{4}^{-, -}$ NaOH ⇌ $N a^{+} + O H^{-}$ 1 mole of $H_{2} S O_{4}$ acid gives 2 moles of $H_{3} O^{+}$ ions. So 2 moles of $O H^{-}$ are required for complete neutralization.

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Equilibrium Question 50

Question: 1×10−12 of 1MH2SO4 will completely neutralise

Options:

Answer:

Solution:

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Question: $1 \times 10^{- 12}$ of $1 M H_{2} S O_{4}$ will completely neutralise