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Chapter 6: Problem 58

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 lead(II) ions from \(150.0 \mathrm{mL}\) of \(0.250 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), \(562.5 \mathrm{mL}\) of \(0.100 M \mathrm{Na}_{3}\mathrm{PO}_{4}\) solution is required.

Step by step solution

01

Write down the balanced chemical equation

First, we need to write down the balanced chemical equation for the reaction between sodium phosphate (Na3PO4) and lead(II) nitrate (Pb(NO3)2). The equation is given by: \[2\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} + 3\mathrm{Na}_{3} \mathrm{PO}_{4} \rightarrow 6\mathrm{NaNO}_{3}+ \mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\]
02

Calculate the moles of Pb(NO3)2

Now, we will calculate the number of moles of Pb(NO3)2 using the given volume and molarity. Moles of Pb(NO3)2 = Molarity × Volume Moles of Pb(NO3)2 = \(0.250 M × 0.150 L = 0.0375 mol\)
03

Calculate the moles of Na3PO4 required

Next, we will use the stoichiometry of the balanced chemical equation to determine the number of moles of Na3PO4 needed to react with the moles of Pb(NO3)2 we calculated in step 2. According to the balanced chemical equation, 3 moles of Na3PO4 react with 2 moles of Pb(NO3)2. Moles of Na3PO4 required = \(\frac{3}{2} \times \text{moles of Pb(NO} _3)_2\) Moles of Na3PO4 required = \(\frac{3}{2} \times 0.0375 mol = 0.05625 mol\)
04

Calculate the volume of Na3PO4 solution needed

Finally, we will use the molarity of the given Na3PO4 solution and the number of moles calculated in step 3 to find the volume of Na3PO4 solution required to precipitate all the lead(II) ions. Volume = Moles ÷ Molarity Volume = \(\frac{0.05625 mol}{0.100 M} = 0.5625 L\) Since the question asked for the volume in milliliters, we multiply by 1000: Volume = \(0.5625 L \times 1000 = 562.5 \mathrm{mL}\) Hence, \(562.5 \mathrm{mL}\) of \(0.100 M \mathrm{Na}_{3}\mathrm{PO}_{4}\) solution is required to precipitate all the lead(II) ions from \(150.0 \mathrm{mL}\) of \(0.250 \mathrm{M}\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\).

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Most popular questions from this chapter

Three students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a 0.1472 -g sample of the salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate had been filtered and dried, it weighed 0.2327 g. Each student analyzed the data independently and came to different conclusions. Pat decided that the metal was titanium. Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt? Look for information on the sulfates of gallium, sodium, and titanium in this text and reference books such as the \(C R C\) Handbook of Chemistry and Physics. What further tests would you suggest to determine which student is most likely correct?

A solution is prepared by dissolving \(10.8 \mathrm{g}\) ammonium sulfate in enough water to make \(100.0 \mathrm{mL}\) of stock solution. A \(10.00-\) mL sample of this stock solution is added to \(50.00 \mathrm{mL}\) of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

A student titrates an unknown amount of potassium hydrogen phthalate \(\left(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4},\) often abbreviated KHP) with \right. \(20.46 \mathrm{mL}\) of a \(0.1000-M \mathrm{NaOH}\) solution. KHP (molar mass \(=\) \(204.22 \mathrm{g} / \mathrm{mol}\) ) has one acidic hydrogen. What mass of KHP was titrated (reacted completely) by the sodium hydroxide solution?

What volume of \(0.0521 M B a(O H)_{2}\) is required to neutralize exactly \(14.20 \mathrm{mL}\) of \(0.141 M \mathrm{H}_{3} \mathrm{PO}_{4} ?\) Phosphoric acid contains three acidic hydrogens.

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) A \(0.205-\mathrm{g}\) sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The \(\mathrm{BaSO}_{4}\) that is formed is filtered, dried, and weighed. Its mass is 0.298 g. What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?
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