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Chapter 4: Problem 45

Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodium bicarbonate.

Short Answer

Expert verified
The concentration of sodium ions in the final solution after mixing 70.0 mL of 3.0 M sodium carbonate and 30.0 mL of 1.0 M sodium bicarbonate is 4.5 M.

Step by step solution

01

Identify the chemical formulas of the compounds

The chemical formula for sodium carbonate is Na2CO3 and for sodium bicarbonate, it is NaHCO3. Notice that sodium carbonate has two sodium atoms per molecule, which can form separate Sodium ions, while sodium bicarbonate has one sodium atom per molecule.
02

Calculate the moles of sodium ions in each solution separately

To calculate the moles of sodium ions in each solution, use the formula: Moles = Volume (L) × Concentration (M) For sodium carbonate: Moles_Na2CO3 = 70.0 mL × (3.0 mol/L) × (1 L / 1000 mL) = 0.21 mol Since there are two sodium ions per molecule, the moles of sodium ions will be: Moles_Na+ = 2 × Moles_Na2CO3 =2 × 0.21 mol = 0.42 mol For sodium bicarbonate: Moles_NaHCO3 = 30.0 mL × (1.0 mol/L) × (1 L / 1000 mL) = 0.03 mol Since there is only one sodium ion per molecule, the moles of sodium ions will be the same as the moles of sodium bicarbonate: Moles_Na+ = Moles_NaHCO3 = 0.03 mol
03

Add the moles of sodium ions together

Now that we have the moles of sodium ions from each solution, we add them together to get the total moles of sodium ions: Total moles_Na+ = Moles_Na+ (from Na2CO3) + Moles_Na+ (from NaHCO3) = 0.42 mol + 0.03 mol = 0.45 mol
04

Calculate the final volume of the mixed solution

To calculate the final volume, simply add the volume of each solution: Final volume = Volume_Na2CO3 + Volume_NaHCO3 = 70.0 mL + 30.0 mL = 100 mL Convert the final volume to liters: Final volume = 100 mL × (1 L / 1000 mL) = 0.1 L
05

Calculate the concentration of sodium ions in the final solution

To find the concentration, divide the total moles of sodium ions by the final volume: Concentration_Na+ = Total moles_Na+ / Final volume = 0.45 mol / 0.1 L = 4.5 M So, the concentration of sodium ions in the final solution is 4.5 M.

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