J.R. S. answered 10/15/19
Ph.D. University Professor with 10+ years Tutoring Experience
2KHC8H4O4(aq) + Ba(OH)2(aq) ---> BaC8H4O4(s) + K2C8H4O4(aq) + 2H2O(l) ... balanced equation
moles KHP = 4.33 g KHP x 1 mol/204.23 g = 0.0212 moles KHP
moles Ba(OH)2 needed = 0.0212 moles KHP x 1 mol Ba(OH)2/2 moles KHP = 0.0106 moles
volume Ba(OH)2 needed: 0.415 mol/L (x L) = 0.0106 moles and x = 0.0255 L = 25.5 mls barium hydroxide
moles KHP = 5.57 g x 1 mol/204.23 g = 0.0273 moles
moles Ba(OH)2 = 0.0273 moles KHP x 1 mole Ba(OH)2/2 moles KHP = 0.0136 moles Ba(OH)2
Concentration of barium hydroxide = 0.0136 moles/0.0326 L = 0.418 M barium hydroxide