Answer
There are necessary 0.250 L of that $Pb(NO_3)_2$ solution to precipitate all the lead(II) ions from that other solution.
Work Step by Step
- Write the reaction between $Na_3PO_4$ and $Pb(NO_3)_2$:
$Na_3PO_4(aq) + Pb(NO_3)_2(aq) -- \gt NaNO_3 + Pb_3(PO_4)_2$
- Balance the reaction. Begin with the number of lead (Pb) atoms:
$Na_3PO_4(aq) + 3Pb(NO_3)_2(aq) -- \gt NaNO_3 + Pb_3(PO_4)_2$
- Balance the number of $P$ atoms:
$2Na_3PO_4(aq) + 3Pb(NO_3)_2(aq) -- \gt NaNO_3 + Pb_3(PO_4)_2$
- Balance the $Na$:
$2Na_3PO_4(aq) + 3Pb(NO_3)_2(aq) -- \gt 6NaNO_3 + Pb_3(PO_4)_2$
- The equation is balanced.
1. According to the balanced equation, 2 moles of $Na_3PO_4$ reacts with 3 moles of $Pb(NO_3)_2$.
2. 1000 mL = 1 L
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Now, use these informations and the data on the exercise as conversion factors to find the $Na_3PO_4$ required volume.
$150.0mL \times \frac{1L}{1000mL} \times \frac{0.250mol(Pb(NO_3)_2)}{1L} \times \frac{2mol(Na_3PO_4)}{3mol(Pb(NO_3)_2)} \times \frac{1L (Solution(Na_3PO_4))}{0.100mol(Na_3PO_4)} = 0.250L(Solution(Na_3PO_4))$
