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2 years ago

Determine the molarity of the sodium ions when 78.0 g Na2S is dissolved in water for a final volume of 1.0 L.

Chemistry

1 answer:

olga55 [171]2 years ago

5 0

Answer:

[Na⁺] = 1.99 M

Explanation:

Na₂S is a ionic salt that can be dissociated.

Dissociation equation is:

Na₂S → 2Na⁺ + S⁻²

1 mol of sodium sulfide can give 2 moles of sodium cation.

We convert moles of salt: 78 g . 1mol / 78.06 g = 0.999 moles

As ratio is 1:2, after dissociation we have (0.999 . 2) = 1.998 moles of Na⁺

Molarity is a type of concentration.

It indicates moles of solute in 1 L of solution and in this case, we have 1 L as final voulme.

Moles of Na⁺ are 1.998 moles. Then molarity (mol/L) is:

M =1.99 mol/L

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As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3

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Hence, the correct option is, (D) 100 times the original content.

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