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Chapter 4: Problem 56

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 \(0.250 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) ?

Short Answer

Expert verified
To precipitate all the lead(II) ions from 150.0 mL of 0.250 M \(Pb(NO_{3})_{2}\) solution, 562.5 mL of 0.100 M \(Na_{3}PO_{4}\) solution is required.

Step by step solution

01

Write the balanced chemical equation for the precipitation reaction

The reaction between lead(II) nitrate (Pb(NO₃)₂) and sodium phosphate (Na₃PO₄) can be written as: \(Pb(NO_{3})_{2} + Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3}\) However, we need to balance this equation. The balanced chemical equation is: \(2Pb(NO_{3})_{2} + 3Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3}\)
02

Calculate moles of lead(II) nitrate

The moles of lead(II) nitrate can be calculated using the formula: Moles = Molarity × Volume in Liters Moles of \(Pb(NO_{3})_{2}\) = 0.250 M × (150.0 mL × (1 L / 1000 mL)) Moles of \(Pb(NO_{3})_{2}\) = 0.0375 mol
03

Determine moles of sodium phosphate required using stoichiometry

From the balanced equation, we can see that 2 moles of lead(II) nitrate react with 3 moles of sodium phosphate. Thus, the mole ratio of \(Pb(NO_{3})_{2}\) to \(Na_{3}PO_{4}\) is 2:3. Using this ratio, we can calculate the required moles of sodium phosphate to completely react with the moles of lead(II) nitrate that we have: Moles of \(Na_{3}PO_{4}\) = (3 moles \(Na_{3}PO_{4}\) / 2 moles \(Pb(NO_{3})_{2}\)) × 0.0375 moles \(Pb(NO_{3})_{2}\) Moles of \(Na_{3}PO_{4}\) = 0.05625 mol
04

Calculate the volume of sodium phosphate solution required

We know the molarity of sodium phosphate solution is 0.100 M, and we have found the moles of sodium phosphate required to react completely with the given amount of lead(II) nitrate. Now we can calculate the required volume of sodium phosphate solution using the formula: Volume in Liters = Moles / Molarity Volume of \(Na_{3}PO_{4}\) = 0.05625 mol / 0.100 M Volume of \(Na_{3}PO_{4}\) = 0.5625 L To convert this result to milliliters, multiply by 1000: Volume of \(Na_{3}PO_{4}\) = 0.5625 L × 1000 mL/L Volume of \(Na_{3}PO_{4}\) = 562.5 mL So, 562.5 mL of 0.100 M sodium phosphate solution is required to precipitate all the lead(II) ions from 150.0 mL of 0.250 M lead(II) nitrate solution.

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Most popular questions from this chapter

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$ \begin{array}{l} \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \\ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) \end{array} $$ A \(10.00-\mathrm{g}\) mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with \(156 \mathrm{~mL}\) of \(3.00 M\) silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If \(78.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) was added, what was the concentration of the \(\mathrm{HCl}\) ?

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\) level can be determined by titrating 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 balanced equation is \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow\) $$ 4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) $$ This reaction is an oxidation-reduction reaction. What species is reduced, and what species is oxidized? How many electrons are transferred in the balanced equation above?

Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only carbon, hydrogen, and oxygen. It was commonly used as a dye in the first half of the nineteenth century. It is \(53.66 \% \mathrm{C}\) and \(4.09 \% \mathrm{H}\) by mass. A titration required \(18.02 \mathrm{~mL}\) of \(0.0406 \mathrm{M} \mathrm{NaOH}\) to neutralize \(0.3602 \mathrm{~g}\) carminic acid. Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{FeSO}_{4}(a q)+\mathrm{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

The drawings below represent aqueous solutions. Solution A is \(2.00 \mathrm{~L}\) of a \(2.00 \mathrm{M}\) aqueous solution of copper(II) nitrate. Solution \(\mathrm{B}\) is \(2.00 \mathrm{~L}\) of a \(3.00 \mathrm{M}\) aqueous solution of potassium hydroxide. a. Draw a picture of the solution made by mixing solutions \(\mathrm{A}\) and \(\mathrm{B}\) together after the precipitation reaction takes place. Make sure this picture shows the correct relative volume compared to solutions \(\mathrm{A}\) and \(\mathrm{B}\), and the correct relative number of ions, along with the correct relative amount of solid formed. b. Determine the concentrations (in \(M\) ) of all ions left in solution (from part a) and the mass of solid formed.
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