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 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.