Question

If 2.00 mL of 0.600 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?

Answer 1

Answer

Balance reaction is

2NaOH + CaCl₂ = Ca(OH)₂ + 2NaCl

1. Calculate number of mole for NaOH and CaCl2

Number of mole of NaOH =molarity x volume in litre = 0.600 x 2.0 x 10^-3           

Number of mole of NaOH = 0.0012 moles

Number of mole of CaCl2 = 0.800 x 1.0 = 0.80 moles

Now total volume = 1.0 L + 2.0 ml = 1.002 L

1. So calculate molarity for NaOH and CaCl2 by using total volume

Molarity of NaOH = number of mole /volume in litre = 0.0012 / 1.002 =0.001198 i.e. = 0.0012 M

Similarly

Molarity of CaCl2 = 0.800 / 1.002 =0.001198 i.e. = 0.7984 M

So reaction Quotient for above reaction ( Q )= [Ca2+][OH-]^2

Q = 0.7984 x (0.0012 )^2

Reaction Quotients Q = 1.15 x 10^-6

Ksp for Ca(OH)₂ is = 5.5 x 10^-6

So Q> Ksp value

So precipitation is occur