Solid potassium chromate is slowly added to 175 mL of a lead (II) nitrate solution until the concentration of chromate ion is 0.0366 M. The maximum amount of lead ion remaining in solution is

[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.

Explanation:

K₂CrO₄ (aq) + Pb (NO₃) ₂ (aq) = > 2KNO₃ (aq) + PbCrO₄ (s)

PbCrO₄ (s) ⇄ Pb⁺² (aq) + CrO₄⁻² (aq)

The K₂CrO₄ (aq) + Pb (NO₃) ₂ (aq) is 1:1 = > moles K₂CrO₄ added = moles of Pb (NO₃) ₂ converted to PbCrO₄ (s) in 175 ml aqueous solution.

Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which means that moles Pb (NO₃) ₂ converted into PbCrO₄ is also 0.0336 mol/L.

Assuming all PbCrO₄ formed in the 175 ml solution remained ionized, then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.

To test if saturation occurs, Qsp must be > than Ksp ...

=> Qsp = [Pb⁺² (aq) ][CrO₄⁻² (aq) ] = (0.0.192) ² = 0.037 >> Ksp (PbCrO₄) = 3 x 10⁻13 = > This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation. That is ...

PbCrO₄ (s) ⇄ Pb⁺² (aq) + CrO₄⁻² (aq); Ksp = 3 x 10⁻¹³

C (i) - - - 0.00M 0.0336M

ΔC - - - + x + x

C (eq) - - - x 0.0336 + x ≅ 0.0336M

Ksp = [Pb⁺² (aq) ][CrO₄⁻² (aq) ] = [Pb⁺² (aq) ] (0.0336M) = 3 x 10⁻¹³

∴[Pb⁺² (aq) ] = (3 x 10⁻¹³/0.0336) M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.

[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.

Explanation:

K₂CrO₄ (aq) + Pb (NO₃) ₂ (aq) = > 2KNO₃ (aq) + PbCrO₄ (s)

PbCrO₄ (s) ⇄ Pb⁺² (aq) + CrO₄⁻² (aq)

The K₂CrO₄ (aq) + Pb (NO₃) ₂ (aq) is 1:1 = > moles K₂CrO₄ added = moles of Pb (NO₃) ₂ converted to PbCrO₄ (s) in 175 ml aqueous solution.

Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which means that moles Pb (NO₃) ₂ converted into PbCrO₄ is also 0.0336 mol/L.

By assumption if we say, all PbCrO₄ formed in the 175 ml solution remained ionized,

Then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.

To test if saturation occurs,

Qsp must be > than Ksp ...

=> Qsp = [Pb⁺² (aq) ][CrO₄⁻² (aq) ] = (0.0.192) ² = 0.037 >> Ksp (PbCrO₄) = 3 x 10⁻13 = >

This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation.

i. e ...

PbCrO₄ (s) ⇄ Pb⁺² (aq) + CrO₄⁻² (aq); Ksp = 3 x 10⁻¹³

C (i) - - - 0.00M 0.0336M

ΔC - - - + x + x

C (eq) - - - x

0.0336 + x ≅ 0.0336M

Ksp = [Pb⁺² (aq) ][CrO₄⁻² (aq) ] = [Pb⁺² (aq) ] (0.0336M) = 3 x 10⁻¹³

Therefore;

[Pb⁺² (aq) ] = (3 x 10⁻¹³/0.0336) M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.