15.0 mL of 0.30 M sodium phosphate solution reacts with 20.0 mL of 0.22 M lead(II) nitrate solution. What is the concentration of phosphate ions left in solution after the reaction is complete?
A) 0.0 M
B) 0.045 M
C) 0.13 M
D) 0.060 M
E) 0.12 M
To find the concentration of phosphate ions left in the solution after the reaction is complete, we need to determine the limiting reactant first and then use the stoichiometry of the balanced equation.
First, let's write the balanced equation for the reaction between sodium phosphate and lead(II) nitrate:
2Na3PO4 + 3Pb(NO3)2 -> 6NaNO3 + Pb3(PO4)2
From the balanced equation, we can see that the ratio of sodium phosphate (Na3PO4) to phosphate ions (PO4) is 2:3.
Next, let's calculate the moles of sodium phosphate and lead(II) nitrate in the solutions:
Moles of sodium phosphate = volume (in L) x concentration (in mol/L)
= 0.015 L x 0.3 mol/L
= 0.0045 mol
Moles of lead(II) nitrate = volume (in L) x concentration (in mol/L)
= 0.020 L x 0.22 mol/L
= 0.0044 mol
Based on the stoichiometry of the balanced equation, we can see that it takes 2 moles of sodium phosphate to react with 3 moles of lead(II) nitrate.
Now, we need to determine the limiting reactant. The reactant that is completely consumed is the limiting reactant. In this case, the limiting reactant is lead(II) nitrate because we have fewer moles of it than sodium phosphate.
Since 2 moles of sodium phosphate react with 3 moles of lead(II) nitrate, and we have 0.0044 mol of lead(II) nitrate, we can calculate the moles of phosphate ions reacted:
Moles of phosphate ions reacted = (3/2) x moles of lead(II) nitrate
= (3/2) x 0.0044 mol
= 0.0066 mol
Finally, we can subtract the moles of phosphate ions reacted from the initial moles of phosphate ions (0.0045 mol) to find the moles of phosphate ions remaining:
Moles of phosphate ions remaining = Initial moles - Moles reacted
= 0.0045 mol - 0.0066 mol
= -0.0021 mol
Since the number of moles cannot be negative, it means that all the phosphate ions have reacted. Therefore, the concentration of phosphate ions left in the solution after the reaction is complete is 0.0 M. So the correct answer is:
A) 0.0 M
To find the concentration of phosphate ions left in solution after the reaction is complete, we need to determine the limiting reactant.
First, let's write the balanced chemical equation for the reaction between sodium phosphate and lead(II) nitrate:
2 Na3PO4 + 3 Pb(NO3)2 → 6 NaNO3 + Pb3(PO4)2
From the balanced equation, we can see that the molar ratio between sodium phosphate and lead(II) nitrate is 2:3. This means that for every 2 moles of sodium phosphate, we need 3 moles of lead(II) nitrate to react completely.
Next, let's calculate the number of moles of sodium phosphate and lead(II) nitrate used in the reaction:
Moles of sodium phosphate = volume (in L) × concentration (in mol/L)
= 0.015 L × 0.30 mol/L
= 0.0045 mol
Moles of lead(II) nitrate = volume (in L) × concentration (in mol/L)
= 0.020 L × 0.22 mol/L
= 0.0044 mol
Since the molar ratio between sodium phosphate and lead(II) nitrate is 2:3, we can see that 0.0044 mol of lead(II) nitrate is the limiting reactant, as it is slightly smaller than 0.0045 mol of sodium phosphate.
Now, let's calculate the number of moles of phosphate ions reacted:
Moles of phosphate ions reacted = 0.0044 mol × (2 mol phosphate ions / 3 mol lead(II) nitrate)
= 0.0029 mol
Finally, let's calculate the concentration of phosphate ions left in solution:
Concentration of phosphate ions left = Moles of phosphate ions left / volume of solution (in L)
= (0.0045 mol - 0.0029 mol) / (0.015 L + 0.020 L)
= 0.0016 mol / 0.035 L
= 0.046 M
Therefore, the concentration of phosphate ions left in solution after the reaction is complete is approximately 0.046 M.
The answer closest to 0.046 M is option D) 0.060 M.
Write and balance the equation.
moles Na3PO4 = M x L
moles Pb(NO3)2 = M x L
Determine which reagent is in excess, from that you go to the ppt formed and from that the amounts of reagents not reacted.
Post your work if you get stuck.