Question

Solubility and Precipitation When 10.00 mL of 1.00 M NaOH are added to 1.00 L of 0.100 M CaCl2 will a precipitate occur? Show you calculations. This is not a True/False question: Just answering “yes” or “no” is not a correct or acceptable answer.

Answer 1

moles of NaOH initially present = 1*10/1000= 0.01 M

Moles of CaCl2 initially present = 0.1*1 =0.1

After mixing , the volume becomes =1+0.01=1.01 L

Concentration of NaOH ( source of OH-) =0.01/1.01 =0.009901 M

NaOH--à Na+ + OH-

Concentration of CaCl2 ( sources of Ca)= 0.1/1.01 =0.09901 M

CaCl2 --- Ca+2 + 2Cl-

Ca(OH)2 --- Ca +2 + 2OH-

Ksp = [Ca+2] [OH-]² =0.09901* [ 0.009901]² =9.7*10^-6

Solubility product ( from literature ) = 5.5*10-6

Hence precipitation takes place