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Conductivity MP BP and Electronegativity (Group 2)

Pb2+ , N2 , B...

Question

$P b^{2 +}, N_{2}, B, P H_{3}, A g, I^{-}$
Name the compound formed from $P b^{2 +}$ and $I^{-}$ ions and write a formula for it.

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Solution

The given ions are $P b^{2 +}$ and $I^{-}$ . Here we need $2 I^{-}$ ions to balance $+ 2$ charge of $P b^{2 +}$ . Therefore compound formed from their combination is $P b I_{2}$ or Lead di Iodide.

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Q. From the given list of cations and anions, write atleast 10 compounds each with its formula and name:

Cation	Anions
$N H_{4}^{+}$	$C l O_{3}^{-}$
$P b^{2 +}$	$S O_{3}^{2 -}$
$F e^{3 +}$	$C r O_{4}^{-}$
$K^{+}$	$H S O_{4}^{-}$

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{A g}^{+}

{P b}^{2 +}

ions are preciptated together by the addition of:

Q. In equilibrium analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains

A g^{+}

P b^{2 +}

at a concentration of

0.10

M. Aqueous HCl is added to this solution until the

C l^{-}

concentration is

0.10

M. What will the concentrations of

A g^{+}

P b^{2 +}

be at equilibrium?
(

K_{s p}

A g C l = 1.8 \times 10^{- 10}

K_{s p}

P b C l_{2} = 1.7 \times 10^{- 5}

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