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nikdorinn [45]
3 years ago
14

What is the Na+ concentration in each of the following solutions:

Chemistry
2 answers:
KIM [24]3 years ago
7 0
<h3>a) <u>Answer;</u></h3>

A- 7.30 M

<h3><u>Explanation;</u></h3>

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.

<em><u>Concentration of sodium ions = 3.65 × 2= 7.3 M</u></em>

<h3>b) <u> Answer;</u></h3>

B- 2.76 M

<h3><u>Explanation;</u></h3>

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.

<em>Concentration of sodium ions = 1.38 ×2 </em>

<em>                                                   = 2.76 M</em>

<h3>c) <u>Answer;</u></h3>

<em>0.785 M</em>

<h3><u>Explanation;</u></h3>

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; <em><u>Na+ concentration = 0.785 M</u></em>

olya-2409 [2.1K]3 years ago
4 0
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
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