Question

11. If the molar concentration of Pbl, is
1.5x10-mol L-?, then the concentration of iodide
ions in g ion L-1 is:-
(1) 3.0 x 10-3
(2) 6.0 x 10-3
(3) 0.3 x 10-3
(4) 0.6 x 10-6​

Answer 1

Explanation:

The correct answer is $\text { (1) } 3.0 \times 10^{-3}$

The explanation is as follows:

Lead iodide is $\mathrm{Pbl}_{2}$

In aqueous medium, $\mathrm{Pbl}_{2}$ dissociates to form

$\mathrm{Pbl}_{2} \rightarrow\left(\mathrm{Pb}^{2+}\right)+2\left(\mathrm{I}^{-}\right)$

Thus, according to the conditions of equilibrium,

$\left[\mathrm{Pbl}_{2}\right] \rightarrow\left[\mathrm{Pb}^{2+}\right]+\left[\mathrm{I}^{-}\right]^{2}$

At equilibrium,  

Concentration of $P b^{2+}=\left[P b^{2+}\right]=1.5 \times 10^{-3} \mathrm{M}$  

Concentration of $\mathrm{I}^{-}=\left[\mathrm{I}^{-}\right]=2 \times 1.5 \times 10^{-3} \mathrm{M}$          

$=3.0 \times 10^{-3} \mathrm{M}$

Thus, If the molar concentration of Pbl, is $1.5 \times 10-3 \mathrm{mol} \mathrm{L}-?$

then the concentration of iodide ions in g ion $\mathrm{L}-1 \text { is }(1) 3.0 \times 10^{-3}$

Answer 2

Answer:

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