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Byju's Answer

Standard X

Chemistry

Zinc Ion

Brown ppt. A ...

Question

Brown ppt. (A) dissolve in $H N O_{3}$ gives (B) which gives white ppt. (C) with $N H_{4} O H$ . (C) on reaction with $H C l$ gives solution (D) which gives white turbidity on addition of water. What is (D) ?

B i C l_{3}

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B i (O H)_{3}

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B i O C l

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B i (N O_{3})_{3}

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Solution

The correct option is A $B i C l_{3}$
$B i_{2} S_{3} + 8 H N O_{3} h o t - - \to 2 B i (N O_{3})_{3} + 3 S ↓ + 2 N O + 4 H_{2} O$
(dil.)
$B i (N O_{3})_{2} + 3 N H_{4} O H \to B i (O H)_{3} ↓ + 3 N H_{4} N O_{3}$
(C)
$B i (O H)_{3} ↓ + 3 H C l \to B i C l_{3} + 3 H_{2} O$
(D)
$B i C l_{3} + H_{2} O ⇌ B i O C l + 2 H C l$
(D) white turbidity

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Similar questions

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(A)

neither gives ppt. with

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(ii)

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H N O_{3}

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(D)

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. Aqueous solution of

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on addition of water gives a whitish turbidity

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(A)

on warming with alkaline sodium stannite gives a black precipitate of a metal

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C

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Brown ppt. (A) dissolve in HNO3 gives (B) which gives white ppt. (C) with NH4OH. (C) on reaction with HCl gives solution (D) which gives white turbidity on addition of water. What is (D) ?

The correct option is A BiCl3Bi2S3+8HNO3hot−−→2Bi(NO3)3+3S↓+2NO+4H2O (dil.)Bi(NO3)2+3NH4OH→Bi(OH)3↓+3NH4NO3 (C)Bi(OH)3↓+3HCl→BiCl3+3H2O (D) BiCl3+H2O⇌BiOCl+2HCl (D) white turbidity

Brown ppt. (A) dissolve in $H N O_{3}$ gives (B) which gives white ppt. (C) with $N H_{4} O H$ . (C) on reaction with $H C l$ gives solution (D) which gives white turbidity on addition of water. What is (D) ?