Question

How many grams of lead(II) nitrate must be dissolved in 1.00 L of water to produce a solution that is 0.300 M in total dissolved ions?

Answer 1

Answer:

A solution that is 0.100 M (i.e. 33,12 g of lead (II) nitrate in 1.00 L of water) yields the desired total dissolved ions concentration.

Explanation:

The molecular formula of lead (II) nitrate is Pb(NO$Pb(NO_{3} )_{2}$ and its molecular mass is 331,2 g/mol.

In disolution, the equilibrium will look like this:

Pb(NO$Pb(NO_{3} )_{2}$ -> $Pb^{2+} + 2(NO)_{3} ^{-1}$

The equation above means that, one mol of lead (II) nitrate dissolved in 1L will yield one mol of Pb ions and 2 moles of NO3 ions, i.e. 3 moles total.

If we dissolve 0.100 moles of lead (II) nitrate in 1.00 L of water, the stoichiometry of the disolution states that in turn, it will yield 0.100 of Pb ions and 0.200 moles of NO3 ions, i.e. 0.300 M in total dissolved ions.

331,2 g/mol * 0.100 mol/L * 1 L = 33,12 grams of the compound are required.