The question is incomplete, the complete question is;
Suppose 0.377g of potassium acetate is dissolved in 250.mL of a 57.0mM aqueous solution of ammonium sulfate.
Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the potassium acetate is dissolved in it.
Round your answer to 3 significant digits.
Answer:
0.0152 M
Explanation:
The equation of the reaction is;
2CH3COOK(aq) + (NH4)2SO4(aq)------> K2SO4(aq) + 2CH3COONH4(aq)
Number of moles of potassium acetate = 0.377g/98.15 g/mol = 0.0038 moles
Number of moles of ammonium sulphate = 250/1000L × 57 × 10^-3 = 0.014 moles
2 moles of potassium acetate yields 2 moles of ammonium acetate
Hence;
0.0038 moles of potassium acetate yields 0.0038 moles of ammonium acetate
Also
1 mole of potassium sulphate yields 2 moles of ammonium acetate
0.014 of potassium sulphate yields 0.014 × 2/1 = 0.028 moles of ammonium acetate
So potassium acetate is the limiting reactant.
Since 0.0038 moles of ammonium acetate is produced, the final concentration of potassium acetate is = 0.0038 moles of ammonium acetate/0.25L = 0.0152 M
Hence final concentration of acetate ions =0.0152 M
