Question

2. What is the total concentration of all ions (the sum of their individual concentrations) in aqueous solutions of the following?  

• Hint: Each substance is an electrolyte that dissociates in water to produce ions. Use the link table of polyatomic ions if you do not recognize the polyatomic ion. For example, Na₂SO₄ -> 2Na¹⁺ + SO₄^2- when it dissolves, so a 1 M soln of Na₂SO₄ is 2M in Na¹⁺ and 1M in SO₄^2- and the total concentration of ions is 2 M + 1 M = 3M.

0.0500-M CaCl2	......M
0.0840-M Na3PO4	......M
0.0220-M K2Cr2O7	......M

Answer 1

PART 1

Consider dissociation of CaCl ₂ in water . CaCl ₂ (aq) $\rightarrow$ Ca ²⁺ (aq) + 2 Cl ^- (aq)

From reaction, 1 mol CaCl ₂$\equiv$ 1 mol Ca ²⁺$\equiv$ 2 mol Cl ^-

$\therefore$ [Ca ²⁺] =[CaCl ₂] = 0.0500 M & [Cl ^-] = 2 [CaCl ₂] = 2 ( 0.0500 M) = 0.100 M

Hence, Total concentration of ions is 0.0500 M + 0.100 M = 0.150 M

PART 2

Consider dissociation of Na₃PO₄ in water . Na₃PO₄ (aq) $\rightarrow$ 3 Na ⁺ (aq) + PO₄ ^3- (aq)

From reaction, 1 mol Na₃PO₄$\equiv$ 3 mol Na ⁺$\equiv$ 1 mol PO₄ ^3-

$\therefore$ [Na ⁺] = 3 [Na₃PO₄] = 3 (0.0840 M ) = 0.252 M & [ PO₄ ^3- ] = [Na₃PO₄] = 0.0840 M

Hence, Total concentration of ions is 0.252 M + 0.0840 M =0.336 M

PART 3

Consider dissociation of K₂Cr₂O₇ in water . K₂Cr₂O₇ (aq) $\rightarrow$ 2 K ⁺ (aq) + Cr₂O₇ ^2- (aq)

From reaction, 1 mol K₂Cr₂O₇ $\equiv$ 2 mol K ⁺$\equiv$ 1 mol Cr₂O₇ ^2-

$\therefore$ [K ⁺] = 2 [K₂Cr₂O₇] = 2 ( 0.0220 M) = 0.0440 M & [ Cr₂O₇ ^2- ] = [K₂Cr₂O₇] = 0.0220 M

Hence, Total concentration of ions is 0.0440 M + 0.0220 M =0.0660 M

ANSWER :

0.0500 M CaCl 2	0.150 M
0.0840 M Na3PO4	0.336 M
0.0220 M K2Cr2O7	0.0660 M