Questions: Which of the following solutions contains the largest number of moles of dissolved ions? a. 50. mL of 1.0 M hydrochloric acid b. 100. mL of 0.5 M nitric acid c. 25 mL of 5.0 M sodium chloride d. 25 mL of 2.0 M sulfuric acid e. 200. mL of 0.10 M sodium hydroxide

Solution Steps

To determine which solution contains the largest number of moles of dissolved ions, we need to calculate the moles of ions for each solution. The number of moles of ions is determined by the concentration of the solution and the dissociation of the compound in water.

1. Hydrochloric acid (HCl):

   • Volume = 50 mL = 0.050 L
   • Molarity = 1.0 M
   • HCl dissociates into H\(^+\) and Cl\(^-\), giving 2 ions per formula unit.
   • Moles of HCl = \(0.050 \, \text{L} \times 1.0 \, \text{mol/L} = 0.050 \, \text{mol}\)
   • Moles of ions = \(0.050 \, \text{mol} \times 2 = 0.100 \, \text{mol ions}\)

2. Nitric acid (HNO\(_3\)):

   • Volume = 100 mL = 0.100 L
   • Molarity = 0.5 M
   • HNO\(_3\) dissociates into H\(^+\) and NO\(_3^-\), giving 2 ions per formula unit.
   • Moles of HNO\(_3\) = \(0.100 \, \text{L} \times 0.5 \, \text{mol/L} = 0.050 \, \text{mol}\)
   • Moles of ions = \(0.050 \, \text{mol} \times 2 = 0.100 \, \text{mol ions}\)

3. Sodium chloride (NaCl):

   • Volume = 25 mL = 0.025 L
   • Molarity = 5.0 M
   • NaCl dissociates into Na\(^+\) and Cl\(^-\), giving 2 ions per formula unit.
   • Moles of NaCl = \(0.025 \, \text{L} \times 5.0 \, \text{mol/L} = 0.125 \, \text{mol}\)
   • Moles of ions = \(0.125 \, \text{mol} \times 2 = 0.250 \, \text{mol ions}\)

4. Sulfuric acid (H\(_2\)SO\(_4\)):

   • Volume = 25 mL = 0.025 L
   • Molarity = 2.0 M
   • H\(_2\)SO\(_4\) dissociates into 2 H\(^+\) and SO\(_4^{2-}\), giving 3 ions per formula unit.
   • Moles of H\(_2\)SO\(_4\) = \(0.025 \, \text{L} \times 2.0 \, \text{mol/L} = 0.050 \, \text{mol}\)
   • Moles of ions = \(0.050 \, \text{mol} \times 3 = 0.150 \, \text{mol ions}\)

5. Sodium hydroxide (NaOH):

   • Volume = 200 mL = 0.200 L
   • Molarity = 0.10 M
   • NaOH dissociates into Na\(^+\) and OH\(^-\), giving 2 ions per formula unit.
   • Moles of NaOH = \(0.200 \, \text{L} \times 0.10 \, \text{mol/L} = 0.020 \, \text{mol}\)
   • Moles of ions = \(0.020 \, \text{mol} \times 2 = 0.040 \, \text{mol ions}\)

Now, we compare the moles of ions calculated for each solution:

• Hydrochloric acid: 0.100 mol ions
• Nitric acid: 0.100 mol ions
• Sodium chloride: 0.250 mol ions
• Sulfuric acid: 0.150 mol ions
• Sodium hydroxide: 0.040 mol ions

Final Answer