Question

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 1

moles of potassium acetate = 0.377 / 98.15

                                            = 3.84 x 10^-3

moles of (NH4)2SO4 = 0.250 x 57 x 10^-3

                                    = 0.01425

(NH4)2SO4 + 2 CH3COOK   -----------------------------> K2SO4 + 2 CH3COONH4

it is double dispalcement reaction so initial concentration of the reactant is same as final concentration .

so

concenttration of potassium acetate = 3.84 x 10^-3 / 0.250

                                                           = 0.0154 M

final molarity of acetate anion = 0.0154 M