1 - For the insoluble salt lead (ii) phosphate Pb3(PO4)2, the solubility product, Ksp is 3.10x10-44...

1 - For the insoluble salt lead (ii) phosphate Pb₃(PO₄)₂, the solubility product, Ksp is 3.10x10^-44

a- Write the chemical equation and solubility product expression for Pb₃(PO₄)₂

b- Determine the molar solubility (M) of Pb₃(PO₄)₂

c- Identify wheather the molar solubility of Pb₃(PO₄)₂ will increase or decrease upon the following changes and explain why?

*** The ddition of 0.10 M Pb(NO₃)₂

*** The addition of 0.10 M NaCl

Solutions

Expert Solution

Given that Ksp of Pb₃(PO₄)₂ = 3.10 x 10^-44

Pb₃(PO₄)₂ ---------------> 3 Pb²⁺ + 2 PO₄^3-

3s 2s s = molar solubility of Pb₃(PO₄)₂

Ksp = [Pb²⁺]³ [PO₄^3-]²

3.10 x 10^-44 = (3s)³ (2s)²

3.10 x 10^-44 = 108 s⁵

s = [3.10 x 10^-44 / 108]^1/5

= 0.78 x 10^-9 M

Therefore, molar solubility of Pb₃(PO₄)₂ = 0.78 x 10^-9 M

c) Molar solubility (M) of Pb₃(PO₄)₂ after the addition of 0.10 M Pb(NO₃)₂

Dut to common ion Pb2+ in Pb(NO₃)₂ and Pb₃(PO₄)_{2 ,} molar solubility of Pb₃(PO₄)₂ decreases.

Molar solubility (M) of Pb₃(PO₄)₂ after the addition of 0.10 M NaCl

Since there is no common ion, molar solubility of Pb₃(PO₄)₂ is not effected.

queen_honey_blossom answered 1 year ago

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