The solubility product of PbI_(2) is 8.0 x 10^(-9). The solubility of lead iodide in 0.1 molar solution of lead nitrate is x x 10^(-6). mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given sqrt 2 = 1.41]

Given, [K_sp]_Pbl_2 = 8 x 10^ - 9To calculate solubility of Pbl_(2) in 0.1 M solution of Pb(NO_(3))_(2),(I) mathop Pb(NO_3)_2limits_0. 1 M -> mathop Pb^2 + (aq)limits_0.1 M + mathop 2NO_3^ - (aq)limits_0. 2 M (II) PbI_2(s) mathop Pb^2 + (aq)limits_S + mathop 2I^ - (aq)limits_2S Therefore [Pb^(2+)] = S + 0.1 = 0.1Because S

The solubility product of PbI2 is 80 x 109 The solubili... | null (Chemistry)

Given,

{[{K_{sp}}]_{Pb{l_2}}} = 8 \times {10^{ - 9}}

To calculate solubility of Pbl₂ in 0.1 M solution of Pb(NO₃)₂,

(I)

\mathop {Pb{{(N{O_3})}_2}}\limits_{0. 1 M} \to \mathop {P{b^{2 + }}(aq)}\limits_{0.1 M} + \mathop {2NO_3^ - (aq)}\limits_{0. 2 M}

(II)

Pb{I_2}(s)

\rightleftharpoons

\mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2{I^ - }(aq)}\limits_{2S}

\therefore

[Pb²⁺] = S + 0.1

\approx

0.1

\because

S < < 0.1

Now, K_sp = 8

\times

10^{$-$
9}

[Pb²] [I^$-$]² = 8

\times

10^{$-$
9}

0.1

\times

(2S)² = 8

\times

10^{$-$
9}

4S² = 8

\times

10^{$-$
8}

\Rightarrow

S = 141

\times

10^{$-$
6} M

\therefore

x = 141

Explanation

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