Respuesta :
Reaction for precipitation is
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.
Precipitation is the process of formation of insoluble solute or the precipitate. 300 mL of 0.100 M sodium chloride must be added to precipitate lead ions.
What is the dilution factor?
The dilution factor is the relation between the initial molarity and the volume with the final molarity and the volume. The formula is given as,
[tex]\rm M_{1}V_{1} = \rm M_{2}{V_{2}[/tex]
The balanced chemical reaction for precipitation is shown as,
[tex]\rm Pb(NO_{3})_{2}(aq) + 2NaCl(aq) \rightarrow PbCl_{2}(s)+2NaNO_{3}(aq)[/tex]
Given,
Molarity of sodium chloride = 0.100 M
Volume sodium chloride = ?
Molarity of lead nitrate = 0.200 M
Volume lead nitrate = 75.0 mL
The volume is calculated as:
[tex]\begin{aligned} \rm M(NaCl) \times V(NaCl) &= \rm 2 \times M(Pb(NO_{3})_{2})\times V (Pb(NO_{3})_{2}) \\\\0.100 \times \rm V &= 2 \times 0.200 \times 75\\\\&= 300 \;\rm mL\end{aligned}[/tex]
Therefore, 300.0 mL of sodium chloride is required.
Learn more about the dilution factor here:
https://brainly.com/question/2186408
