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Chapter 4: Problem 39

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 in the resulting mixture is \(4.5 M\).

Step by step solution

01

Determine moles of sodium ions in each solution

We can calculate the moles of sodium ions in each solution by using the formula: moles = volume (in L) × concentration (in M). For the sodium carbonate solution: Volume = \(70.0 mL = 0.070 L\) Concentration = 3.0 M Since sodium carbonate (\(Na_2CO_3\)) contains 2 moles of sodium ions per mole of compound: Moles of sodium ions = \(0.070 L \times 3.0 M \times 2 = 0.42\, moles\) For the sodium bicarbonate solution: Volume = \(30.0 mL = 0.030 L\) Concentration = 1.0 M Since sodium bicarbonate (\(NaHCO_3\)) contains 1 mole of sodium ions per mole of compound: Moles of sodium ions = \(0.030 L \times 1.0 M \times 1 = 0.03\, moles\)
02

Determine total moles of sodium ions and total volume

Now that we have determined the moles of sodium ions from both the sodium carbonate and the sodium bicarbonate solutions, we combine the moles of sodium ions from both solutions to get the total moles. Total moles of sodium ions = \(0.42\, moles + 0.03\, moles = 0.45\, moles\) Next, we need to find the total volume of the mixed solution: Total volume = volume of sodium carbonate solution + volume of sodium bicarbonate solution Total volume = \(0.070 L + 0.030 L = 0.100 L\)
03

Calculate the concentration of sodium ions

Now that we know the total moles of sodium ions and the total volume of the solution, we can find the concentration of sodium ions in the resulting mixture by dividing the total moles by the total volume. The formula for concentration is: \(concentration = \frac{moles}{volume}\). Concentration of sodium ions = \(\frac{0.45\, moles}{0.100 L} = 4.5 M\) The sodium ion concentration in the resulting mixture is 4.5 M.

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Most popular questions from this chapter

Consider the reaction of \(19.0 \mathrm{~g}\) of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and \(29.0 \mathrm{~g}\) of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture.

You are given a \(1.50-\mathrm{g}\) mixture of sodium nitrate and sodium chloride. You dissolve this mixture into \(100 \mathrm{~mL}\) of water and then add an excess of \(0.500 \mathrm{M}\) silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of \(0.641 \mathrm{~g}\). a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the percent sodium chloride in the original unknown mixture.

Consider an experiment in which two burets, \(\mathrm{Y}\) and \(\mathrm{Z}\), are simultaneously draining into a beaker that initially contained \(275.0 \mathrm{~mL}\) of \(0.300 \mathrm{M} \mathrm{HCl}\). Buret \(\mathrm{Y}\) contains \(0.150 \mathrm{M} \mathrm{NaOH}\) and buret \(Z\) contains \(0.250 \mathrm{M} \mathrm{KOH}\). The stoichiometric point in the titration is reached \(60.65\) minutes after \(\bar{Y}\) and \(Z\) were started simultaneously. The total volume in the beaker at the stoichiometric point is \(655 \mathrm{~mL}\). Calculate the flow rates of burets \(\mathrm{Y}\) and \(\mathrm{Z}\). Assume the flow rates remain constant during the experiment.

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

A solution was prepared by mixing \(50.00 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{HNO}_{3}\) and \(100.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the molarity of the final solution of nitric acid.
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