Question

The ksp for BaF2 is 2.4x10^-5. When 52.0 mL of 0.055 M NaF is mixed with 38.0 mL of 0.015 M Ba(NO3)2, will a precipitate form? Rationalize.

Answer 1

oo Condition of precipitation Consider ionic product (KIP) similar to reaction quotient Q in chemical equilibrium. For precipitation, ionic product (KIP) should be greater than solubility product Ksp. This will make the equilibrium of undissolved salt and dissolved salt shift in backward direction leading to precipitation. After precipitation, solution will become saturated and Kip = Ksp. Remember to modify the concentration of the precipitating ions because of volume change occouring upon mixing both solutions. oo

Bafels) = Bat'any + 2 fra) -9.4*10-5 gonic preoduct on fit . - [Ba2+] [Fog? Reaction Qoutient (Q1 = [Bezét! We have, 152mL of 0.055 M Naf solu - Moles of Naf - Molaritylm) x Volume = 0.055 X 52 29.86 mmol NaF — Ng @ + F ©jaqj 2.86 mmol 3.86mmol 2.86 mmol so moles of so = 2.86 mmol 0 38 mL of Moles of 0.015 M Bq (NO3/2 golu Ba|NO3)2 = molarity (M) X Volum = 0.015 * 38 = 0.57 mmol

Ba(NO3), - B92+ + 2NO3- 0.57mmola 0 57 mmol 2x0. 57 mmol so, Moles of Ba2+ = 9.14 mmol total volume of final solution = (52+ 38mL = gomi E so, concentration of F@-[F"]= Moly Volume = 1.86 mmol 90 m2 = 0.03178 M [Ba2+] = 1.14 namol go na = 0.01267 m so, Q = [8q+2] [F-12

Q = [0.01267 ] (0.0317872 Q = 1.98 %10-5 Since - Q<kop so, precipitation will not form after mixing.