Question

If 200.0 mL of 2.5 x 10-4M Pb(NO3)2 is mixed with 300.0 mL of

3.0 x 10-3M CaCl2, will a precipitate form?

Answer 1

No precipitate will form when 200.0 mL of 2.5 x
`10^-4M`
`Pb(NO_3)_2` is mixed with 300.0 mL of 3.0 x
`10^-3M`
`CaCl_2`.

To determine if a precipitate will form when mixing 200.0 mL of 2.5 x
`10^-4M`
`Pb(NO_3)_2` with 300.0 mL of 3.0 x
`10^-3M`
`CaCl_2`, we need to check if a reaction will occur between the two solutions. If a reaction does occur, a precipitate may form.

First, let's write the balanced chemical equation for the reaction between
`Pb(NO_3)_2` and
`CaCl_2`:


`Pb(NO_3)_2` +
`2CaCl_2`→
`PbCl_2`+
`2Ca(NO_3)_2`

Now, let's calculate the moles of each compound in the solution:

Moles of
`Pb(NO_3)_2` = concentration x volume
= (2.5 x
`10^-4` mol/L) x (0.2000 L)
= 5.0 x
`10^-5` mol

Moles of
`CaCl_2`= concentration x volume
= (3.0 x
`10^-3`mol/L) x (0.3000 L)
= 9.0 x
`10^-4`mol

Based on the balanced chemical equation, we can see that 1 mole of
`Pb(NO_3)_2` reacts with 2 moles of
`CaCl_2`. Since the moles of
`Pb(NO_3)_2` (5.0 x
`10^-5`mol) is much less than the moles of
`CaCl_2`(9.0 x
`10^-4M`mol),
`Pb(NO_3)_2` is the limiting reactant.

Since
`Pb(NO_3)_2` is the limiting reactant, it will completely react with the available
`CaCl_2`. This means that all of the
`Pb(NO_3)_2` will be used up in the reaction, and no excess
`Pb(NO_3)_2` will be left to form a precipitate.

Therefore, no precipitate will form when 200.0 mL of 2.5 x
`10^-4M`
`Pb(NO_3)_2` is mixed with 300.0 mL of 3.0 x
`10^-3M`
`CaCl_2`.