Question

What is the Na+ concentration in each of the following solutions:
(A) 3.65 M Sodium Sulfate=
(B) 1.38 M Sodium Carbonate=
(C) 0.785 M Sodium Bicarbonate=

Answer 1

The give compounds are examples of salts which means that these compounds completely dissociates in water. For example in a, sodium sulfate dissociates into 2 moles of Na+ and 1 mole of sulfate ion. In this case, the Na+ concentration should be twice the original conc. Thus, the answer is 7.30 M. In B, the answer is 2.76 M while C has an answer of 0.785 M

Answer 2

a) Answer;

A- 7.30 M

Explanation;

3.65 Sodium sulfate

Na2(SO4) dissociates to give sodium ions and sulfate ions.

Na2(SO4) → 2Na+ + SO₄²₋

Therefore, twice the concentration of sodium ions as far as molarity is concerned.

Concentration of sodium ions = 3.65 × 2= 7.3 M

b) Answer;

B- 2.76 M

Explanation;

b) 1.38 M sodium carbonate

Sodium carbonate dissociates completely to yield Sodium ions and carbonate ions

Na₂CO₃ → 2Na+ + CO₃²₋

The concentration of sodium ions will be twice the concentration of initial compound since it has a ratio of two.

Concentration of sodium ions = 1.38 ×2

                                                  = 2.76 M

c) Answer;

0.785 M

Explanation;

b) 0.785 sodium bicarbonate

Sodium bicarbonate dissociates completely to sodium ions and a bicarbonate ions.

NaHCO₃ →Na+ + HCO₃⁻

In this case the concentration of Na+ will be equal to the concentration of the original compound since they share the same ratio.

Thus; Na+ concentration = 0.785 M