Respuesta :
The IP is greater than Ksp. Lead (II) chloride precipitate will develop. The assertion is accurate.
If a compound's component ion concentration rises over the maximum that its saturated solution can support, precipitation will result.
Take a look at a binary compound:
A + B ⇆ A⁺ + B⁻
This compound's ionic product will be:
IP = [A⁺] [B⁻]
This product's value for precipitation should be higher than its value for solubility. This is,
IP ≥ Ksp
Strong electrolyte lead nitrate entirely dissolves in water as:
Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻
Therefore, we can write
[Pb²⁺] = [Pb(NO₃)₂]
⇒ [Pb²⁺] = 0.12 M
To find the chloride ion concentration:
No. of moles of NaCl (n) = Given mass/ Molar mass
⇒ n = 140/58.5 g/mol = 0.14/58.4 g/mol = 0.00239 moles
NaCl concentration will be:
[NaCl] = n/V = 0.00239/0.25 = 0.00956 M
Strong electrolyte NaCl totally dissociates as follows in solution:
NaCL (aq) → Na⁺ (aq) + Cl⁻ (aq)
Therefore, [Cl⁻] = [NaCl]
⇒ [Cl⁻] = 0.009 M
The lead chloride solubility equilibrium is now provided by:
PbCl₂ ⇄ Pb²⁺ + 2 Cl⁻
As a result, lead (II) chloride's ionic product (IP) in solution will be:
IP = [Pb²⁺] [Cl⁻]² = 0.12 X 0.01024 = 0.0012288
The solubility product of this compound is :
Ksp = 1.7 X 10⁻⁵
Therefore, IP > Ksp
As a result, lead (II) chloride precipitate will develop. The assertion is accurate.
Learn more about the Solubility equilibrium with the help of the given link:
https://brainly.com/question/13719354
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