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.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-

it's c (engineer's do their work with mechanical stuff not on any living being)