Question

2.8. Calculate the concentration of barium in the solution at equilibrium when 15.0mL of 0.200M K2CrO4 is added to 25.0mL of 0.100 M BaCl2. K for BaCrO4 is 2.4 100

Answer 1

The balance reaction of K2 CrO4 and BaCl2 is as follows: K2 CrO4 + BaCl2 -----> BaCrO4(s) + 2 KC1 Given that 15.0 ml of 0.200 M K2 Cr204 25.0 ML OF 0.100 m BaC12 Number of moles = amount in g* molar mass (15.0 mL) x (0.200 M K2CrO4) = 3.00 mmol K2 CrO4 (25.0 mL) x (0.100 M BaCl2)= 2.50 mmol BaCl2 BaCl2 is the limiting reactant The limiting agent has following properties: I.It completely reacted in the reaction. II.It determines the amount of the product in mole. Mole of BaCrO4 = 2.50 mmol BaC12* 1.00 mole BaCr04/1.00 mol BaC12 = 2.50 mmol BaCr04 Total volume =(15.0 mL +25.0 mL) = 40.0 ml=0.040 L Molarity = number of mole / volume in L = 2.50*10^-3 mole s/ 0.040 L =0.0625 M, Given that Ksp=2.4*10^-10 Ksp = [Ba2++] [CrO4--] 2.4*10^-10 = [X+0.0625 ] [X] Here x is very small as compare 0.0625 So 2.4*10^-10 = [0.0625 ] [X] X = 3.84*10^-9