296 g pb(no3)4 and 98.3 g pbcl4 are added to 90 l water. what are [pb4+], [no3-], and [cl-]?

1) [Pb⁴⁺] = 0.01 mol/L.

2) [NO³⁻] = 0.0288 mol/L.

3) [Cl⁻] = 0.01244 mol/L.

Explanation:

Firstly, we need to calculate the concentration of the mentioned salts (Pb(NO₃)₄ and PbCl₄) in water.

Molarity = (n) solute/ V solution,

n of solute is the no. of moles of solute.

V is the volume of the solution in L (V = 90.0 L).

We can calculate the no. of moles of each salt using the relation:

n = mass/molar mass.

∴ no. of moles of (Pb(NO₃)₄) = mass/molar mass = (296.0 g)/(455.22 g/mol) = 0.65 M.

∴ no. of moles of (PbCl₄) = mass/molar mass = (98.3 g)/(349.012 g/mol) = 0.28 M.

∴ M of (Pb(NO₃)₄) = [no. of moles of (Pb(NO₃)₄)] / V of the solution = [0.65 M]/[90.0 L] = 7.22 x 10⁻³ mol/L.

∴ M of (PbCl₄) = [no. of moles of (PbCl₄)] / V of the solution = [0.28 M]/[90.0 L] = 3.11 x 10⁻³ mol/L.

Now, we can calculate the concentration of the different ions  [Pb⁴⁺], [NO³⁻], and [Cl⁻]:

1) [Pb⁴⁺]:

The concentration of  [Pb⁴⁺] will come from the dissociation of the two salts (Pb(NO₃)₄) and (PbCl₄):

Pb(NO₃)₄ → Pb⁴⁺ + 4NO₃⁻,

Every 1.0 mole of (Pb(NO₃)₄) will give 1.0 mole of [Pb⁴⁺ ] ions.

PbCl₄ → Pb⁴⁺ + 4Cl⁻.

Every 1.0 mole of (PbCl₄) will give 1.0 mole of [Pb⁴⁺ ] ions.

∴ [Pb⁴⁺] = [Pb⁴⁺] of (Pb(NO₃)₄) + [Pb⁴⁺] of (PbCl₄)] = (7.22 x 10⁻³ mol/L) + (3.11 x 10⁻³ mol/L) = 0.01 mol/L.

2) [NO³⁻]:

The concentration of  [NO³⁻] will come only from the dissociation of the (Pb(NO₃)₄):

Pb(NO₃)₄ → Pb⁴⁺ + 4NO₃⁻.

Every 1.0 mole of (Pb(NO₃)₄) will give 4.0 mole of [NO³⁻] ions.

∴ [NO³⁻] = 4 x concentration of (Pb(NO₃)₄) = 4 x (7.22 x 10⁻³ mol/L) = 0.0288 mol/L.

3) [Cl⁻]:

The concentration of  [Cl⁻] will come only from the dissociation of the (PbCl₄):

PbCl₄ → Pb⁴⁺ + 4Cl⁻.

Every 1.0 mole of (PbCl₄) will give 4.0 mole of [Cl⁻] ions.

∴ [Cl⁻] = 4 x concentration of (PbCl₄) = 4 x (3.11 x 10⁻³ mol/L) = 0.01244 mol/L.

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