Question

Predict if a mixture of Ba 2+ (20 mL of aqueous 0.02 solution as the nitrate salt) and NaOH (1 mL of aqueous 3 M solution) will form a precipitate. Provide a mathematical explanation in terms of Ksp, and Qsp.  

Answer 1

Ba+2 solution = 20 mL of 0.02 M

NaOH solution = 1 mL of 3M

Total volume = 20 + 1 = 21 mL

after mixing,

Concentration of Ba+2 ions = 0.02M x 20 mL / 21 mL = 0.0190M [ M1V1 = M2V2, M2 = M1V1 / V2 ]

[Ba+2] = 0.0190M

Concentration of NaOH = 3 M x 1 mL / 21 mL = 0.143 M

[OH-] = 0.143 M

Ba(OH)2(s) --------------------Ba+2(aq) + 2 OH-(aq)

Qsp = [ Ba+2] [ OH-]^2

Qsp = 0.0190 x ( 0.143)^2

Qsp= 0.000388

Qsp = 3.88 x10^-4

Ksp of Ba(OH)2 = 5 x10^-3

Qsp < Ksp

Ionic product of the Ions is less than the Solubility product of Ba(OH)2

Hence, No, precipitite will form.