Question

Determine the chloride ion concentration in each of the following solutions:

0.130 M BaCl₂:

M

0.666 M NaCl:

M

1.802 M AlCl₃:

M

(b) What is the concentration of a Sr(NO₃)₂ solution that is 1.55 M in nitrate ion?

M

Answer 1

# 0.130 M BaCl₂

BaCl₂ dissociate as

BaCl₂    Ba²⁺  + 2 Cl^-

1 mole of BaCl₂ produce 2 mole of Cl^- ion therefore concentration of chlorine ion in 0.130 M BaCl₂ =

0.130 2 = 0.260 M

chlorine ion concentration = 0.260 M

# 0.666 M NaCl

NaCl dissociate as

NaCl   Na⁺  + Cl^-

1 mole of NaCl produce 1 mole of Cl^- ion therefore concentration of chlorine ion in 0.666 M NaCl = 0.666 M

chlorine ion concentration = 0.666 M

# 1.802 M AlCl₃

AlCl₃ dissociate as

AlCl₃   Al³⁺  + 3 Cl^-

1 mole of AlCl₃ produce 3 mole of Cl^- ion therefore concentration of chlorine ion in 1.802 M AlCl₃ =

1.802 3 = 5.406 M

chlorine ion concentration = 5.406 M

(b)

Sr(NO₃)₂ dissociate as

Sr(NO₃)₂   Sr²⁺  + 2 NO₃^-

1 mole of Sr(NO₃)₂ produce 2 mole of NO₃^- ion therefore concentration of Sr(NO3)₂ =

1.55 / 2 = 0.775 M

Sr(NO₃)₂ concentration = 0.775 M