Statement 1: $H_{2} S O_{4}$ acts as a base in $H C l O_{4}$ .
Statement 2: Perchloric acid is a stronger acid than $H_{2} S O_{4}$ .

Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation of Statement 1

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Statement 1 is True, Statement 2 is True, Statement 2 is not a correct explanation of Statement 1

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Statement 1 is true, Statement 2 is False

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Statement 1 is False, Statement 2 is True

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Solution

The correct option is A Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation of Statement 1
Perchloric acid is a stronger acid than sulfuric acid. Hence, perchloric acid can protonate sulfuric acid. Hence, sulfuric acid acts as a base in perchloric acid.
$H C l O_{4} \to H^{+} + C l O_{4}$

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

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

$H C l O_{4} + H_{2} S O_{4} ⇋ C l O_{4}^{-} + H_{3} S O_{4}^{+}$

Thus, Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation of Statement 1.

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Similar questions

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

n \geq 2,

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n}

\sqrt{n (n + 1)} < n + 1

N H_{3}

P H_{3}

N H_{3}

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