Suppose you add 0.2469 g of PbCl2(s) to 50.0 mL of water. In the resulting saturated solution, you find that the concentration of PbX2+(aq) is 0.0159 M and the concentration of C l − ( a q ) ClX−(aq) is 0.0318 M. What is the value of the equilibrium constant, Ksp, for the dissolution of PbCl2?

6.431 x 10∧-5 {mol/l)³

Explanation:

The Ksp (PbCl₂) of dissolve PbCl₂ is 6.431×10⁻⁵ (mol/L)³

Explanation:

                      PbCl₂ (s)   ⇒   Pb²⁺ (aq) + 2Cl⁻ (aq)

=           a                      -                 -

=    a - s                 s                2s

T=time , a=initial condition , s=solubility on concentration

Con. [Pb²⁺(aq)] = 0.0159 M

Con. [Cl⁻(aq)] = 0.0318 M

Ksp = [Pb²⁺] [Cl⁻]²

      = (S) (2S)²

      = (0.0159 M) (2 × 0.0318)²

Ksp (PbCl₂) = 6.431×10⁻⁵ M³

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

5.6⋅10−^19  j

do you need the explaination?