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Salt of Weak Acid and Strong Base

The correct r...

Question

The correct relationship between $p H$ of isomolar solution of sodium oxide ( $p H_{1}$ ), sodium sulphide ( $p H_{2}$ ), sodium selenide ( $p H_{3}$ ) and sodium telluride ( $p H_{4}$ ) is:

p H_{1} > p H_{2} > p H_{3} > p H_{4}

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p H_{1} < p H_{2} < p H_{3} \leq p H_{4}

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p H_{1} < p H_{2} < p H_{3} < p H_{4}

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p H_{1} > p H_{2} ≃ p H_{3} > p H_{4}

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Solution

The correct option is A $p H_{1} > p H_{2} > p H_{3} > p H_{4}$
The compounds sodium oxide, sodium sulphide, sodium selenide and sodium telluride respectively will form the following:

$N a O H, N a O H + H_{2} S, N a O H + H_{2} S e$ and $N a O H + H_{2} T e$ on dissolution.

strong base, strong base + weak acid, strong base+ acid, base+strong acid

Acid strength:
$H_{2} S < H_{2} S e < H_{2} T e$

So , the order will be $p H_{1} > p H_{2} > p H_{3} > p H_{4}$

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(p H_{1}),

(p H_{2}),

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(p H_{1})

(p H_{2})

(p H_{3})

(p H_{4})

p H = (p H)_{1}

p H = (p H)_{2}

(p H)_{2} - (p H)_{1} =

(p H)_{1} + (p H)_{2} = 9.5 u n i t s

P H_{3} + \oplus N H_{4} ⟶ \oplus P H_{4} + N H_{3}

N H_{3} + P {\oplus H}_{4} ⟶ N {\oplus H}_{4} + P H_{3}

(C H_{3})_{3} P + \oplus N H_{4} ⟶ (C H_{3})_{3} \oplus P H + N H_{3}

(C H_{3})_{3} N + P H_{4}^{\oplus} ⟶ (C H_{3})_{3} \oplus N H + P H_{3}

P H_{3}, P H_{4}^{+},

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