Final answer:
The final molarity of sodium cation in the solution when 8.46 g of sodium acetate is dissolved in 200 mL of a 0.40 M aqueous solution of ammonium sulfate is 0.515 M, rounded to three significant digits.
Step-by-step explanation:
To calculate the final molarity of sodium cation in the solution when 8.46 g of sodium acetate is dissolved in 200. mL of a 0.40 M aqueous solution of ammonium sulfate, we must consider that sodium acetate dissociates completely into sodium ions (Na+) and acetate ions (CH₃COO−) due to it being a strong electrolyte. First, the number of moles of sodium acetate can be found using its molar mass (82.03 g/mol for CH₃COONa).
The calculations are as follows:
Number of moles of sodium acetate = mass/molar mass
= 8.46 g / 82.03 g/mol
= 0.1031 mol
Since sodium acetate dissociates into one sodium ion per molecule, the moles of sodium ions will be equal to the moles of sodium acetate. Now, the molarity of sodium ions is calculated by dividing the moles of sodium ions by the volume of the solution in liters (0.200 L).
The molarity of sodium ions in the solution is therefore:
Molarity (Na+) = moles of Na+ / volume in L
= 0.1031 mol / 0.200 L
= 0.515 M
Round this value to three significant digits results in a final concentration of 0.515 M for sodium ions in the solution.