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Hard water often contains dissolved Mg2+and Ca2+ions. Water softeners often use sodium carbonate to soften the water. Sodium carbonate is soluble in water, but the carbonate ions form insoluble precipitates with the calcium and magnesium ions, removing them from solution. Suppose a 2.91L solution is 0.0396 M in calcium nitrate and 0.0661 M in magnesium chloride. What mass of sodium carbonate would have to be added to this solution to completely eliminate the hard water ions?

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jufsanabriasa

Answer:

32.5g of sodium carbonate

Explanation:

Reaction of sodium carbonate (Na₂CO₃) with Mg²⁺ and Ca²⁺ as follows:

Na₂CO₃(aq) + Ca²⁺(aq) → CaCO₃(s)

Na₂CO₃(aq) + Mg²⁺(aq) → MgCO₃(s)

1 mole of carbonate reacts per mole of the cations.

To know the mass of sodium carbonate we must know the moles of carbonate we need to add based on the moles of the cations:

Moles Mg²⁺:

2.91L * (0.0661 moles MgCl₂ / 1L) = 0.192 moles MgCl₂ = Moles Mg²⁺

Moles Ca²⁺:

2.91L * (0.0396mol Ca(NO₃)₂ / 1L) = 0.115 moles Ca(NO₃)₂ = Moles Ca²⁺

That means moles of sodium carbonate you must add are:

0.192 moles + 0.115 moles = 0.307 moles sodium carbonate.

In grams (Using molar mass Na₂CO₃ = 105.99g/mol):

0.307 moles Na₂CO₃ * (105.99g / mol) =

32.5g of sodium carbonate

dsdrajlin

A 2.91 L solution that is 0.0396 M in calcium nitrate and 0.0661 M in magnesium chloride requires 32.5 g of sodium carbonate to completely eliminate the hard water ions.

We have 2.91 L of a solution that is 0.0396 M in calcium nitrate and 0.0661 M in magnesium chloride. We will calculate the number of moles of each metal of the type M²⁺.

[tex]2.91 L \times \frac{0.0396molCa(NO_3)_2}{L} \times \frac{1molCa^{2+} }{1molCa(NO_3)_2} = 0.115 molCa^{2+} \\\\2.91 L \times \frac{0.0661molMgCl_2}{L} \times \frac{1molMg^{2+} }{1molMgCl_2} = 0.192 molMg^{2+}[/tex]

The total number of moles of metals of the type M²⁺ is:

[tex]n = 0.115mol+0.192mol = 0.307 mol[/tex]

Sodium carbonate reacts with the metals of the type M²⁺ according to the following equation.

Na₂CO₃ + M²⁺ ⇒ MCO₃ + 2 Na⁺

We can calculate the mass of Na₂CO₃ needed to react with 0.307 moles of M²⁺ considering the following relationships.

  • The molar ratio of Na₂CO₃ to M²⁺ is 1:1.
  • The molar mass of Na₂CO₃ is 105.99 g/mol.

[tex]0.307molM^{2+} \times \frac{1molNa_2CO_3}{1molM^{2+} } \times \frac{105.99gNa_2CO_3}{1molNa_2CO_3} = 32.5gNa_2CO_3[/tex]

A 2.91 L solution that is 0.0396 M in calcium nitrate and 0.0661 M in magnesium chloride requires 32.5 g of sodium carbonate to completely eliminate the hard water ions.

Learn more: https://brainly.com/question/22288091

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