Question

Be sure to answer all parts. If 2.00 mL of 0.100 M NaOH are added to 1.000 L of 0.800 M CaCl2, what is the value of the reaction quotient and will precipitation occur? Q =   × 10 (Enter your answer in scientific notation.)    Precipitation will occur.    Precipitation will not occur.

Answer 1

no of moles of NaOH = molarity * volume in L

                                   = 0.1*0.002   = 0.0002 moles

   NaOH ----------> Na⁺ + OH^-

                                        0.0002 moles

      [OH-]   = no of moles/volume in L

                 = 0.0002/1.002   = 0.0002M

no of moles of CaCl2 = molarity * volume in L

                                     = 0.8*1 = 0.8moles

CaCl2 -------------> Ca⁺² + 2Cl^-

0.8 moles                0.8 moles

[Ca⁺²]   = no of moles/volume in L

             = 0.8/1.002   = 0.798 M

Qsp = [Ca⁺²][OH^-]²

          = 0.798*(0.0002)² = 3.2*10^-8

ksp of Ca(OH)2 = 5.02*10^-8

Ksp>Qsp will not form precipitate

Precipitation will not occur.