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Chapter 6: Problem 41

Calculate the sodium ion concentration when \(70.0 \mathrm{mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

Short Answer

Expert verified
The sodium ion concentration when \(70.0 \mathrm{mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate is \(4.5\mathrm{M}\).

Step by step solution

01

Convert volume to liters

The volumes of solutions are given in milliliters. We need to convert them into liters, as concentration is typically measured in moles per liter. \(70.0 \mathrm{mL}\) of sodium carbonate solution is equal to \(0.070 \mathrm{L}\) and \(30.0 \mathrm{mL}\) of sodium bicarbonate solution is equal to \(0.030 \mathrm{L}\).
02

Calculate moles of sodium ions in sodium carbonate solution

The formula of sodium carbonate is \(\mathrm{Na_2CO_3}\). Therefore, there are 2 moles of sodium ions for every mole of sodium carbonate. We'll multiply the concentration of sodium carbonate by the volume of the solution in liters and then multiply by the number of moles of sodium ions per mole of sodium carbonate. Moles of sodium ions in sodium carbonate solution = concentration × volume × (number of moles of sodium ions per mole of sodium carbonate) Moles of sodium ions in sodium carbonate = \(3.0\mathrm{M} \times 0.070\mathrm{L} \times 2 = 0.42\mathrm{moles}\)
03

Calculate moles of sodium ions in sodium bicarbonate solution

The formula of sodium bicarbonate is \(\mathrm{NaHCO_3}\). Therefore, there is 1 mole of sodium ions for every mole of sodium bicarbonate. We'll multiply the concentration of sodium bicarbonate by the volume of the solution in liters. Moles of sodium ions in sodium bicarbonate solution = concentration × volume × (number of moles of sodium ions per mole of sodium bicarbonate) Moles of sodium ions in sodium bicarbonate = \(1.0\mathrm{M} \times 0.030\mathrm{L} \times 1 = 0.03\mathrm{moles}\)
04

Calculate total moles of sodium ions

To find the total moles of sodium ions, we'll add the moles of sodium ions from both solutions. Total moles of sodium ions = moles of sodium ions in sodium carbonate + moles of sodium ions in sodium bicarbonate Total moles of sodium ions = \(0.42 + 0.03 = 0.45\mathrm{moles}\)
05

Calculate the total volume of mixed solution

To calculate the total volume of the mixed solution, we'll add the volumes of both solutions. Total volume of mixed solution = volume of sodium carbonate solution + volume of sodium bicarbonate solution Total volume = \(0.070\mathrm{L} + 0.030\mathrm{L} = 0.100\mathrm{L}\)
06

Calculate the sodium ion concentration in the mixed solution

To find the concentration of sodium ions in the mixed solution, we'll divide the total moles of sodium ions by the total volume of the mixed solution. Sodium ion concentration = \(\frac{total\ moles\ of\ sodium\ ions}{total\ volume\ of\ mixed\ solution}\) Sodium ion concentration = \(\frac{0.45}{0.100}\mathrm{M} = 4.5\mathrm{M}\) Therefore, the sodium ion concentration when \(70.0 \mathrm{mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate is \(4.5\mathrm{M}\).

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Most popular questions from this chapter

Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. copper(II) sulfate \((a q)+\) iron \((s) \longrightarrow\) copper \((s)\) # iron(II) sulfate \((a q)\) copper \((\text { II) sulfate }(a q) \text { # iron }(s) \Longrightarrow\) copper \((s)+\) iron(III) sulfate \((a q)\) You place \(87.7 \mathrm{mL}\) of a \(0.500-M\) solution of copper(II) sulfate in a beaker. You then add \(2.00 \mathrm{g}\) of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate 2.27 g of copper. Which equation above describes the reaction that occurred? Support your answer.

When organic compounds containing sulfur are burned, sulfur dioxide is produced. The amount of \(\mathrm{SO}_{2}\) formed can be determined by the reaction with hydrogen peroxide: $$ \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ The resulting sulfuric acid is then titrated with a standard NaOH solution. A 1.302 -g sample of coal is burned and the \(\mathrm{SO}_{2}\) is collected in a solution of hydrogen peroxide. It took \(28.44 \mathrm{mL}\) of a \(0.1000-M \mathrm{NaOH}\) solution to titrate the resulting sulfuric acid. Calculate the mass percent of sulfur in the coal sample. Sulfuric acid has two acidic hydrogens.

A 30.0 -mL. sample of an unknown strong base is neutralized after the addition of \(12.0 \mathrm{mL}\) of a \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) solution. If the unknown base concentration is 0.0300 \(M,\) give some possible identities for the unknown base.

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid-base reactions. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) b. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q) \rightarrow\)

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of 0.5894 g were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of BaSO_ obtained was \(0.5032 \mathrm{g} .\) What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?
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