The solubility product of Pbl₂ is 8.0 × 10^-⁹. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^-⁶ mol/L. The value of x is…………………… (Rounded off to the nearest integer)

[Given $\sqrt[]{2} = 1.41]$

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Vishal Baghel | Contributor-Level 10

2 months ago

$P b l_{2} ? P b^{2 +} + 2 l^{-} K_{s q} = 8.0 \times 1 0^{- 9}$

Pb (NO₃)₂ -> Pb²⁺ +^{$2 N O_{3}^{-}$}

0.1 M-

-0.1 M 0.1 M

$K_{s q} = 8 \times 1 0^{- 9} U s i n g K_{s q} = [P b^{2 +}] {[I^{-}]}^{2}$

$8 \times 1 0^{- 9} = 0.1 \times {(2 S)}^{2}$

S = 141 × 10^-6 M

<!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816608695"> </o:OLEObject></xml><! [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816608695"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mi>P</mi> <mi>b</mi> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>? </mo> <mi>P</mi> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> <mo>+</mo> </mrow> </msup> <mo>+</mo> <mn>2</mn> <msup> <mrow> <mi>l</mi> </mrow> <mrow> <mo>−</mo> </mrow> </msup> <mtext> </mtext> <msub> <mrow> <mi>K</mi> </mrow> <mrow> <mi>s</mi> <mi>q</mi> </mrow> </msub> <mo>=</mo> <mn>8</mn> <mo>.</mo> <mn>0</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>9</mn> </mrow> </msup> </mrow> </math> Pb (NO3)2 -> Pb2+ + <math> <mrow> <mn>2</mn> <mi>N</mi> <msubsup> <mrow> <mi>O</mi> </mrow> <mrow> <mn>3</mn> </mrow> <mrow> <mo>−</mo> </mrow> </msubsup> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1816608696"> </o:OLEObject></xml><! [endif]->0.1 M--0.1 M    0.1 M  <math> <mrow> <msub> <mrow> <mi>K</mi> </mrow> <mrow> <mi>s</mi> <mi>q</mi> </mrow> </msub> <mo>=</mo> <mn>8</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>9</mn> </mrow> </msup> <mo></mo> <mi>U</mi> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>g</mi> <mtext> </mtext> <mtext> </mtext> <msub> <mrow> <mi>K</mi> </mrow> <mrow> <mi>s</mi> <mi>q</mi> </mrow> </msub> <mo>=</mo> <mrow> <mo> [</mo> <mrow> <mi>P</mi> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> <mo>+</mo> </mrow> </msup> </mrow> <mo>]</mo> </mrow> <msup> <mrow> <mrow> <mo> [</mo> <mrow> <msup> <mrow> <mi>I</mi> </mrow> <mrow> <mo>−</mo> </mrow> </msup> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>         <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1816608697"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mn>8</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>9</mn> </mrow> </msup> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>1</mn> <mo>×</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mi>S</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> S = 141 × 10-6 M

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The solubility product of Pbl₂ is 8.0 × 10^-⁹. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^-⁶ mol/L. The value of x is…………………… (Rounded off to the nearest integer)

[Given $\sqrt[]{2} = 1.41]$

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The solubility product of Pbl2 is 8.0 × 10-9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10-6 mol/L. The value of x is…………………… (Rounded off to the nearest integer) [Given 2 = 1 . 4 1 ]

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The solubility product of Pbl₂ is 8.0 × 10^-⁹. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^-⁶ mol/L. The value of x is…………………… (Rounded off to the nearest integer)

[Given $\sqrt[]{2} = 1.41]$