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Chapter 4: Problem 62

What volume of 0.100\(M \mathrm{Na}_{3} \mathrm{PO}_{4}\) is required to precipitate all the lead(II) ions from 150.0 \(\mathrm{mL}\) of 0.250$M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ?$

Short Answer

Expert verified
We need 375 mL of 0.100 M Na3PO4 solution to precipitate all the lead(II) ions from 150.0 mL of 0.250 M Pb(NO3)2 solution.

Step by step solution

01

Calculate the moles of Pb(NO3)2 present in the solution

First, we need to determine the number of moles of Pb(NO3)2 in 150.0 mL of 0.250 M solution. We can find this by using the formula: Moles = (Molarity) × (Volume in Liters) Convert the volume of the solution from mL to L: Volume in Liters = 150.0 mL × (1 L / 1000 mL) = 0.150 L Now, calculate the moles of Pb(NO3)2: Moles of Pb(NO3)2 = (0.250 M) × (0.150 L) = 0.0375 moles
02

Determine the moles of Na3PO4 required to precipitate all Pb(II) ions

Using the stoichiometry from the balanced equation, we need 1 mole of Na3PO4 for every 1 mole of Pb(NO3)2. So, we require the same number of moles of Na3PO4 as we calculated for Pb(NO3)2. Moles of Na3PO4 required = Moles of Pb(NO3)2 = 0.0375 moles
03

Calculate the volume of 0.100 M Na3PO4 solution needed

Now, we will find the volume of 0.100 M Na3PO4 solution needed to provide the required moles of Na3PO4. We can use the formula: Volume in Liters = (Moles) / (Molarity) Calculate the volume of Na3PO4 solution needed: Volume of Na3PO4 solution = (0.0375 moles) / (0.100 M) = 0.375 L
04

Convert the volume to milliliters

Finally, we will convert the volume of Na3PO4 solution from liters to milliliters: Volume in mL = 0.375 L × (1000 mL / 1 L) = 375 mL Thus, we need 375 mL of 0.100 M Na3PO4 solution to precipitate all the lead(II) ions from 150.0 mL of 0.250 M Pb(NO3)2 solution.

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Most popular questions from this chapter

Assign oxidation states for all atoms in each of the following compounds. a. $\mathrm{UO}_{2}^{2+} \quad \quad f. \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}$ b. $\mathrm{As}_{2} \mathrm{O}_{3} \quad \quad g. \mathrm{Na}_{2} \mathrm{P}_{2} \mathrm{O}_{3}$ c. \(\mathrm{NaBiO}_{3} \quad h. \mathrm{Hg}_{2} \mathrm{Cl}_{2}\) d. $\mathrm{As}_{4} \quad\quad \quad i. \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ e. \(\mathrm{HAsO}_{2}\)

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by tirrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If 31.05 \(\mathrm{mL}\) of 0.0600\(M\) potassium dichromate solution is required to titrate 30.0 \(\mathrm{g}\) of blood plasma, determine the mass percent of alcohol in the blood.

Many plants are poisonous because their stems and leaves contain oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) or sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) . When ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion, $\mathrm{C}_{2} \mathrm{O}_{4}^{2-},$ in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride, \(\mathrm{CaCl}_{2},\) in aqueous solution.

A solution of permanganate is standardized by titration with oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right) .\) It required 28.97 \(\mathrm{mL}\) of the permanganate solution to react completely with 0.1058 g of oxalic acid. The unbalanced equation for the reaction is $$\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \stackrel{\mathrm{Acidic}}{\longrightarrow} \mathrm{Mn}^{2+}(a q)+\mathrm{CO}_{2}(g)$$ What is the molarity of the permanganate solution?

The thallium (present as \(\mathrm{Tl}_{2} \mathrm{SO}_{4} )\) in a \(9.486-\mathrm{g}\) pesticide sam- ple was precipitated as thallium(I) iodide. Calculate the mass percent of \(\mathrm{TI}_{2} \mathrm{SO}_{4}\) in the sample if 0.1824 \(\mathrm{g}\) of TIII was recovered.
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