Question

0.050 M HCl Volume at equivalence point = 10.37 mL

Volume of Ca(OH)2 used in the titraiton = 15 mL

a) What happens at equivalence point?

b) Calculate the [OH-].

c) Calculate the [Ca+2].

d) Calculate the Ksp for Ca(OH)2.

e) What is the concentration (in molarity) of the saturated Ca(OH)2?

Answer 1

Ca(OH)2 + 2HCl -----> CaCl2 + 2H2O

a) at equivalence point the total HCl is neutralised by Ca(OH)2.

b) M1V1 = M2V2

0.05*10.37/15 = 0.034 M = molarity of Ca(OH)2.

concentration of Ca(OH)2 at equivalence point = 15/(15+10.37)*0.034 = 0.02 M

pOH= -log(2*0.02) = 1.4

[OH-] = 10^(-pOH) = 10^(-1.4) = 0.04 M

c) [Ca+2]. = 0.02

d) Ksp = [Ca2+][OH-]

      = 0.02*0.04 = 0.0008

E) What is the concentration (in molarity) of the saturated Ca(OH)2 = 0.02 M