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Chapter 4: Problem 54

A \(5.00 \mathrm{mL}\) sample of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) requires \(49.1 \mathrm{mL}\) of \(0.217 \mathrm{M} \mathrm{NaOH}\) to convert all of the \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) to \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) The other product of the reaction is water. Calculate the molarity of the \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution.

Short Answer

Expert verified
The molarity of the Na\(_{2}\)H\(_{3}\)PO\(_{4}\) solution is 1.065 M.

Step by step solution

01

Write and Balance the Chemical Equation

First, we need to correctly write the chemical equation for the reaction and balance it: \[2 \mathrm{NaOH}+\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Na}_{3} \mathrm{PO}_{4}\] The coefficients tell us that 2 moles of NaOH reacts with 1 mole of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) to produce 2 moles of water and 1 mole of \(\mathrm{Na}_{3} \mathrm{PO}_{4}.\)
02

Calculate the number of moles of NaOH

We can calculate the number of moles of NaOH using its molarity and volume. The molarity (M) is the number of moles of solute per liter of solution, so \[Moles \, of \, NaOH = Molarity \, of \, NaOH \times Volume\] \[Moles \, of \, NaOH = 0.217M \times 49.1mL = 0.0219 \, moles\] We need to convert milliliters to liters by dividing by 1000, so \(Moles \, of \, NaOH = 0.0219 moles \times \frac{1L}{1000mL} = 0.01065 mole\)
03

Apply the stoichiometric relationships

From the balanced chemical equation, it is clear that 2 moles of NaOH react with 1 mole of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\). Therefore, the number of moles of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) in the reaction would be half of the number of moles of NaOH: \[Moles \, of \, \mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4} = \frac{0.01065}{2} = 0.005325 moles\]
04

Calculate the Molarity of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\)

We can calculate the molarity of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) using the formula: \[Molarity = \frac{Moles \, of \, solute}{Volume \, of \, solution} = \frac{0.005325 \, moles}{5.00 \, mL} \] Again, we need to convert milliliters to liters by dividing by 1000: \[Molarity = \frac{0.005325 \, moles}{5.00 \, mL/1000} = 1.065 \, M\] Thus, the molarity of the \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution is 1.065 M.

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Most popular questions from this chapter

In the decomposition of silver(I) carbonate to form metallic silver, carbon dioxide gas, and oxygen gas, (a) one mol of oxygen gas is formed for every 2 mol of carbon dioxide gas; (b) 2 mol of silver metal is formed for every 1 mol of oxygen gas; (c) equal numbers of moles of carbon dioxide and oxygen gases are produced; (d) the same number of moles of silver metal are formed as of the silver(I) carbonate decomposed.

Consider the chemical equation below. What is the maximum number of moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be obtained from a reaction mixture containing 5.0 moles each of \(\mathrm{KMnO}_{4}, \mathrm{KI},\) and \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) \(3.0 \mathrm{mol}\); (b) \(3.8 \mathrm{mol}\) (c) \(5.0 \mathrm{mol} ;\) (d) \(6.0 \mathrm{mol} ;\) (e) \(15 \mathrm{mol}\). $$\begin{array}{r} 2 \mathrm{KMnO}_{4}+10 \mathrm{KI}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 6 \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+5 \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O} \end{array}$$

Solid silver oxide, \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}),\) decomposes at temperatures in excess of \(300^{\circ} \mathrm{C},\) yielding metallic silver and oxygen gas. A 3.13 g sample of impure silver oxide yields \(0.187 \mathrm{g} \mathrm{O}_{2}(\mathrm{g}) .\) What is the mass percent \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample? Assume that \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})\) is the only source of \(\mathrm{O}_{2}(\mathrm{g}) .\) [Hint: Write a balanced equation for the reaction.]

In the reaction of \(277 \mathrm{g} \mathrm{CCl}_{4}\) with an excess of \(\mathrm{HF}\) \(187 \mathrm{g} \mathrm{CCl}_{2} \mathrm{F}_{2}\) is obtained. What are the (a) theoretical, (b) actual, and (c) percent yields of this reaction? $$\mathrm{CCl}_{4}+2 \mathrm{HF} \longrightarrow \mathrm{CCl}_{2} \mathrm{F}_{2}+2 \mathrm{HCl}$$

How many grams of commercial acetic acid (97\% \(\mathrm{CH}_{3} \mathrm{COOH}\) by mass) must be allowed to react with an excess of \(\mathrm{PCl}_{3}\) to produce \(75 \mathrm{g}\) of acetyl chloride \(\left(\mathrm{CH}_{3} \mathrm{COCl}\right),\) if the reaction has a \(78.2 \%\) yield? $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{PCl}_{3} & \longrightarrow \mathrm{CH}_{3} \mathrm{COCl}+ \mathrm{H}_{3} \mathrm{PO}_{3}(\text { not balanced }) \end{aligned}$$
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