Question

Calculate Qsp for calcium fluoride (Ksp = 3.9 ×10–11) when 140.0 mL of a 5.10×10-3 M solution of Ca(NO3)2 is added to 180.0 mL of a 7.70×10-3 M solution of KF.

Answer 1

ak Moscaa) Given, Ca(NO3)3 = 5.10x109 mvolume = 140.0mL = 650 0.1404 KF = 7.70x163 m, volume = 180.0mL = 0.180L Ksp = 3.9x1011 asp =? of cafz Reaction for formation of Catz is- C Ca(NO3) + QKF Cafe + a KNO3 (99) (89) Net equation - cazt + 2F Cafy :: Asp = [c"][F] We need to calculate concentration of cazt & fePos reaction.foom given data. moles of cat - moles Ca(Nex) = molarity x volume in Litter = 5010X163 x 0.140 L moles of cazt = 71.4 Xpeed 105 mol Total volume of solution after adding caruozbz&KF is= 92 cm = 320 mL = 0-3201 molarity of Ca2+ = Flat Stob – 71.40x175 mol molarity of cat = 0.00223 M Same way, moles of F- moles of KF = Molarity x volume in Litter = 7.70x103 mol x 0.1802 moles of F = 0.00139 moles molarity of F = 0.00139 miles (molomity of F = 0.00434 M माना 0.320L 0 320L csscanned with CamScanner

Asp = [cat][F-12 =[0.00223] [0.0043472 Asp = 4.2x108 CScanned with CamScanner
As here Qsp value is greater than Ksp value we can say that precipitate is form on mixing of two solutions.