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Chapter 4: Problem 110

Sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) is available in very pure form and can be used to standardize acid solutions. What is the molarity of a HCl solution if \(28.3 \mathrm{~mL}\) of the solution are required to react with \(0.256 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\)

Short Answer

Expert verified
The molarity of the HCl solution is \(0.170 \, \text{M}\).

Step by step solution

01

Find the balanced chemical equation

Based on the molecular structure of the compounds, the balanced equation between sodium carbonate and hydrochloric acid is:\[\mathrm{Na}_{2} \mathrm{CO}_{3} + 2 \mathrm{HCl} \longrightarrow 2 \mathrm{NaCl} + \mathrm{H}_{2} \mathrm{O} + \mathrm{CO}_{2}\]This reaction shows that for each mole of sodium carbonate, two moles of hydrochloric acid are required.
02

Calculate moles of Na2CO3

First, calculate the amount of moles of sodium carbonate (Na2CO3) that were used. From the periodic table, the molar mass of Na2CO3 is (23.0 * 2) + 12.01 + (16.00 * 3) = 105.99 g/mol.So, the moles of Na2CO3 is given by the formula:\[\text{{moles}} = \frac{{\text{{mass}}}}{{\text{{molar mass}}}} = \frac{{0.256 \, \text{g}}}{{105.99 \, \text{g/mol}}} = 0.00241 \, \text{mol}\]
03

Calculate molarity of HCl

Since there are 2 moles of HCl for every mole of Na2CO3, the moles of HCl that reacted is 2 * 0.00241 mol = 0.00482 mol.The molarity (M) is defined as the number of moles of solute per liter of solution. Therefore, the molarity of the HCl solution can be obtained by dividing the calculated moles of HCl by the volume in liters.\[M = \frac{{\text{{moles of solute}}}}{{\text{{volume of solution in liters}}}} = \frac{{0.00482 \, \text{mol}}}{{28.3 \, \text{mL}} \times \frac{{1 \, \text{L}}}{{1000 \, \text{mL}}}} = 0.170 \, \text{M}\]

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Most popular questions from this chapter

Many proteins contain metal ions for structural and/ or redox functions. Which of the following metals fit into one or both categories: \(\mathrm{Ca}, \mathrm{Cu}, \mathrm{Fe}, \mathrm{Mg}, \mathrm{Mn}\) \(\mathrm{Ni}, \mathrm{Zn} ?\)

Hydrochloric acid is not an oxidizing agent in the sense that sulfuric acid and nitric acid are. Explain why the chloride ion is not a strong oxidizing agent like \(\mathrm{SO}_{4}^{2-}\) and \(\mathrm{NO}_{3}^{-}\).

Describe the laboratory preparation for the following gases: (a) hydrogen, (b) oxygen, (c) carbon dioxide, (d) nitrogen. Indicate the physical states of the reactants and products in each case. [Hint: Nitrogen can be obtained by heating ammonium nitrite \(\left(\mathrm{NH}_{4} \mathrm{NO}_{2}\right)\).]

Calculate the concentration (in molarity) of a \(\mathrm{NaOH}\) solution if \(25.0 \mathrm{~mL}\) of the solution are needed to neutralize \(17.4 \mathrm{~mL}\) of a \(0.312 \mathrm{M} \mathrm{HCl}\) solution.

Calculate the volume in milliliters of a \(1.420 \mathrm{M}\) \(\mathrm{NaOH}\) solution required to titrate the following solutions. (a) \(25.00 \mathrm{~mL}\) of a \(2.430 \mathrm{M} \mathrm{HCl}\) solution (b) \(25.00 \mathrm{~mL}\) of a \(4.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution (c) \(25.00 \mathrm{~mL}\) of a \(1.500 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution
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