Question

(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

(a) 0.15 M Na₂CO₃ has the highest concentration of sodium ions (0.30 M). (b) 20.0 mL of 0.15 M NaHCO₃ solution contains a greater number of moles of sodium ions (0.003 moles) than 15.0 mL of 0.04 M Na₂S solution (0.0012 moles).

Step-by-step solution

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 Na₂CO₃: \( [Na^{+}] = 2 \times c_{Na_{2}CO_{3}}\) For Na₃PO₄: \( [Na^{+}] = 3 \times c_{Na_{3}PO_{4}} \) Where c denotes the molarity of respective compounds and [Na⁺] 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: \( c_{NaCl} = 0.25M \) Na₂CO₃: \( c_{Na_{2}CO_{3}} = 0.15M \) Na₃PO₄: \( c_{Na_{3}PO_{4}} = 0.075M \) For NaCl: \( [Na^{+}] = 1 \times 0.25 M = 0.25 M \) For Na₂CO₃: \( [Na^{+}] = 2 \times 0.15 M = 0.30 M \) For Na₃PO₄: \( [Na^{+}] = 3 \times 0.075 M = 0.225 M \) #Step 3: Compare the sodium ion concentrations to find the highest#

Compare the calculated concentrations of sodium ions to find which solution has the highest concentration: For NaCl: \( 0.25 M \) For Na₂CO₃: \( 0.30 M \) For Na₃PO₄: \( 0.225 M \) Thus, we can conclude that 0.15M Na₂CO₃ 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: NaHCO₃: \( c_{NaHCO_{3}} = 0.15M \) Na₂S: \( c_{Na_{2}S} = 0.04M \) For NaHCO₃: \( n_{NaHCO_{3}} = 1 \times c_{NaHCO_{3}}\) For Na₂S: \( n_{Na_{2}S} = 2 \times c_{Na_{2}S} \) #Step 2: Calculate moles of sodium ions in the given volumes#

Now, let's use the given volumes and convert them to liters, then calculate the moles of sodium ions in each: Volume of NaHCO₃ solution: \(20.0 mL = 0.020 L \) Volume of Na₂S solution: \(15.0 mL = 0.015 L \) For NaHCO₃: \( n_{Na^{+}} = c_{NaHCO_{3}} \times V_{NaHCO_{3}}= 0.15M \times 0.020 L = 0.003 moles \) For Na₂S: \( n_{Na^{+}} = c_{Na_{2}S} \times V_{Na_{2}S}= 2 \times (0.04M \times 0.015 L) = 0.0012 moles \) #Step 3: Compare the moles of sodium ions in both solutions#

Now, let's compare the moles of sodium ions in both solutions: For NaHCO₃: \( 0.003 moles \) For Na₂S: \( 0.0012 moles \) Thus, 20.0 mL of 0.15 M NaHCO₃ solution contains a greater number of moles of sodium ions than 15.0 mL of 0.04 M Na₂S solution.

Key concepts

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 (Na₂CO₃) contains two sodium ions per molecule. Likewise, sodium phosphate (Na₃PO₄) 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, Na₂CO₃, and Na₃PO₄, 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.