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Chapter 4: Problem 69

(a) Which will have the highest concentration of sodium ions: \(0.25 \mathrm{MNaCl}, 0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3},\) or \(0.075 \mathrm{MNa}_{3} \mathrm{PO}_{4} ?(\mathbf{b})\) Which will contain the greater number of moles of sodium ion: \(20.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{NaHCO}_{3}\) or \(15.0 \mathrm{~mL}\) of \(0.04 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S} ?\)

Short Answer

Expert verified
(a) 0.15 M Na鈧侰O鈧� has the highest concentration of sodium ions (0.30 M). (b) 20.0 mL of 0.15 M NaHCO鈧� solution contains a greater number of moles of sodium ions (0.003 moles) than 15.0 mL of 0.04 M Na鈧係 solution (0.0012 moles).

Step by step solution

01

To find the concentration of sodium ions in each solution, we have to consider the number of sodium ions present in each compound's formula. Multiply the compound's molarity by the number of sodium ions present in its formula, like this: For NaCl: \( [Na^{+}] = 1 \times c_{NaCl}\) For Na鈧侰O鈧�: \( [Na^{+}] = 2 \times c_{Na_{2}CO_{3}}\) For Na鈧働O鈧�: \( [Na^{+}] = 3 \times c_{Na_{3}PO_{4}} \) Where c denotes the molarity of respective compounds and [Na鈦篯 denotes the sodium ion concentration. #Step 2: Calculate sodium ion concentration for each solution#

Now, let's calculate the sodium ion concentration for each solution using the given molarities. Given: NaCl: \( c_{NaCl} = 0.25M \) Na鈧侰O鈧�: \( c_{Na_{2}CO_{3}} = 0.15M \) Na鈧働O鈧�: \( c_{Na_{3}PO_{4}} = 0.075M \) For NaCl: \( [Na^{+}] = 1 \times 0.25 M = 0.25 M \) For Na鈧侰O鈧�: \( [Na^{+}] = 2 \times 0.15 M = 0.30 M \) For Na鈧働O鈧�: \( [Na^{+}] = 3 \times 0.075 M = 0.225 M \) #Step 3: Compare the sodium ion concentrations to find the highest#
02

Compare the calculated concentrations of sodium ions to find which solution has the highest concentration: For NaCl: \( 0.25 M \) For Na鈧侰O鈧�: \( 0.30 M \) For Na鈧働O鈧�: \( 0.225 M \) Thus, we can conclude that 0.15M Na鈧侰O鈧� has the highest concentration of sodium ions (0.30 M). #b# #Step 1: Identify the moles of sodium in each solution#

In order to compare the number of moles of sodium ions in the given solutions, we have to consider the number of sodium ions per formula unit and multiply that by the given molarity. Given: NaHCO鈧�: \( c_{NaHCO_{3}} = 0.15M \) Na鈧係: \( c_{Na_{2}S} = 0.04M \) For NaHCO鈧�: \( n_{NaHCO_{3}} = 1 \times c_{NaHCO_{3}}\) For Na鈧係: \( n_{Na_{2}S} = 2 \times c_{Na_{2}S} \) #Step 2: Calculate moles of sodium ions in the given volumes#
03

Now, let's use the given volumes and convert them to liters, then calculate the moles of sodium ions in each: Volume of NaHCO鈧� solution: \(20.0 mL = 0.020 L \) Volume of Na鈧係 solution: \(15.0 mL = 0.015 L \) For NaHCO鈧�: \( n_{Na^{+}} = c_{NaHCO_{3}} \times V_{NaHCO_{3}}= 0.15M \times 0.020 L = 0.003 moles \) For Na鈧係: \( n_{Na^{+}} = c_{Na_{2}S} \times V_{Na_{2}S}= 2 \times (0.04M \times 0.015 L) = 0.0012 moles \) #Step 3: Compare the moles of sodium ions in both solutions#

Now, let's compare the moles of sodium ions in both solutions: For NaHCO鈧�: \( 0.003 moles \) For Na鈧係: \( 0.0012 moles \) Thus, 20.0 mL of 0.15 M NaHCO鈧� solution contains a greater number of moles of sodium ions than 15.0 mL of 0.04 M Na鈧係 solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sodium Ion Concentration
Understanding sodium ion concentration is crucial in chemistry, especially when comparing solutions. The concentration of sodium ions refers to the number of sodium ions per unit volume of solution. It's essential to consider the chemical formula of each compound, as it tells how many sodium ions are present in each formula unit. For instance, in the compound sodium chloride (NaCl), there is one sodium ion, while sodium carbonate (Na鈧侰O鈧�) contains two sodium ions per molecule. Likewise, sodium phosphate (Na鈧働O鈧�) consists of three sodium ions. Therefore, to find the total sodium ion concentration in a solution, you need to multiply the number of sodium ions in the formula by the molarity of the solution. This simple multiplication gives a clear picture of how concentrated the sodium ions are, allowing for easier comparison of different solutions. With this understanding, you can analyze which compound will yield a higher sodium ion concentration given their molarities.
Chemical Formulas
Chemical formulas are an integral part of understanding the composition of compounds. They denote the types and numbers of atoms present in a molecule, often indicating the number of ions per formula unit. For sodium-containing compounds like NaCl, Na鈧侰O鈧�, and Na鈧働O鈧�, the chemical formulas are particularly useful for identifying the number of sodium ions. These formulas allow chemists to calculate the exact concentration of ions in a solution. By identifying how many sodium ions each formula provides, you can multiply this by the molarity of the solution to determine the sodium ion concentration. This method exemplifies the power of chemical formulas in solving stoichiometric calculations and ensuring accurate chemical analysis. Whether you're a student learning about chemical formulas for the first time, or a professional in the field, understanding these representations is key to making quick and informed conclusions about chemical compounds.
Solution Comparison
Comparing chemical solutions involves analyzing various aspects, such as concentration, volume, and the number of moles. When dealing with solutions, you'll often need to determine which has the higher concentration of a specific ion or contains more moles of that ion. This requires understanding both the chemical makeup of the solutes and the context of given concentrations and volumes.
For example, in one scenario, you might be given a set of solutions with known molarities and asked to find out which one has the highest concentration of ions. Here, chemical understanding helps you multiply given molarity by the number of pertinent ions from the formula, providing an effective solution comparison.
  • In another scenario, when comparing the number of moles, you also need to consider the solution's volume.
  • Converting milliliters to liters is significant, as calculations are typically done in liters.
  • Presenting your comparisons using moles and concentrations allows for clear, analytical results.
Solution comparison aids not just in academic exercises but forms the basis for practical chemistry applications, such as creating solutions in a laboratory setting.

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Most popular questions from this chapter

Calicheamicin gamma-1, \(\mathrm{C}_{\mathrm{ss}} \mathrm{H}_{74} \mathrm{IN}_{3} \mathrm{O}_{21} \mathrm{~S}_{4},\) is one of the most potent antibiotics known: one molecule kills one bacterial cell. Describe how you would (carefully!) prepare \(25.00 \mathrm{~mL}\) of an aqueous calicheamicin gamma- 1 solution that could kill \(1.0 \times 10^{8}\) bacteria, starting from a \(5.00 \times 10^{-9} \mathrm{M}\) stock solution of the antibiotic.

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathbf{b}) \mathrm{NO}_{3}^{-}\), (d) \(\mathrm{S}^{2-}\), (c) \(\mathrm{NH}_{4}^{+}\) (e) \(\mathrm{SO}_{4}^{2-} .[\) Section 4.2\(]\)

Hard water contains \(\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}\), and \(\mathrm{Fe}^{2+}\), which interfere with the action of soap and leave an insoluble coating on the insides of containers and pipes when heated. Water softeners replace these ions with \(\mathrm{Na}^{+} .\) Keep in mind that charge balance must be maintained. (a) If \(1500 \mathrm{~L}\) of hard water contains \(0.020 \mathrm{M} \mathrm{Ca}^{2+}\) and \(0.0040 \mathrm{M} \mathrm{Mg}^{2+}\), how many moles of Nat are needed to replace these ions? (b) If the sodium is added to the water softener in the form of \(\mathrm{NaCl}\), how many grams of sodium chloride are needed?

Neurotransmitters are molecules that are released by nerve cells to other cells in our bodies, and are needed for muscle motion, thinking, feeling, and memory. Dopamine is a common neurotransmitter in the human brain. (a) Predict what kind of reaction dopamine is most likely to undergo in water: redox, acid-base, precipitation, or metathesis? Explain your reasoning. (b) Patients with Parkinson's disease suffer from a shortage of dopamine and may need to take it to reduce symptoms. An IV (intravenous fluid) bag is filled with a solution that contains \(400.0 \mathrm{mg}\) dopamine per \(250.0 \mathrm{~mL}\) of solution. What is the concentration of dopamine in the IV bag in units of molarity? (c) Experiments with rats show that if rats are dosed with \(3.0 \mathrm{mg} / \mathrm{kg}\) of cocaine (that is, \(3.0 \mathrm{mg}\) cocaine per \(\mathrm{kg}\) of animal mass), the concentration of dopamine in their brains increases by \(0.75 \mu M\) after 60 seconds. Calculate how many molecules of dopamine would be produced in a rat (average brain volume \(5.00 \mathrm{~mm}^{3}\) ) after 60 seconds of a \(3.0 \mathrm{mg} / \mathrm{kg}\) dose of cocaine.

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}^{3-},\) what mass of sodium arsenate would be present in a 1.00-L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$ \mathrm{Ppb}=\frac{\mathrm{g} \text { solute }}{\mathrm{g} \text { solution }} \times 10^{9} $$
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