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A 0.4000 g sample that contained 96.4% Na2SO4 (142.04 g/mol) required 41.25 mL of barium chloride solution to reach the endpoint in a precipitation titration. What is the molarity of the BaCl2 solution?

Respuesta :

maacastrobr

Answer:

0.0658 M

Explanation:

The reaction that takes place is:

Na₂SO₄(aq) + BaCl₂(aq) ↔ BaSO₄(s) + 2NaCl(aq)

First we calculate the number of Na₂SO₄ moles:

  • 0.4000 g sample * 96.4/100 = 0.3856 g Na₂SO₄
  • 0.3856 g Na₂SO₄ ÷ 142.04 g/mol = 2.715x10⁻³ mol Na₂SO₄

Now we convert to moles of BaCl₂:

  • 2.715x10⁻³ mol Na₂SO₄ * 1 molBaCl₂/1 molNa₂SO₄ = 2.715x10⁻³ mol BaCl₂

Finally we divide by the volume to calculate the molarity:

  • 41.25 mL ⇒ 41.25 / 1000 =  0.04125 L
  • 2.715x10⁻³ mol BaCl₂ / 0.04125 L = 0.0658 M

The molarity of [tex]BaCl_{2}[/tex] solution would be as follows:

[tex]0.0658 M[/tex]

Given that

Reaction:

[tex]Na_{2} SO_{4}(aq) + BaCl_{2}(aq)[/tex] → [tex]BaSO_{4}(s) + 2NaCl(aq)[/tex]

To find the moles of [tex]Na_{2}SO_{4}[/tex],

[tex]= 0.4000g[/tex] × [tex]96.4/100[/tex]

[tex]= 0.3856[/tex]

[tex]0.3856[/tex] ÷ [tex]142.04[/tex]

[tex]= 2.715[/tex] × [tex]10^-3 mol[/tex] [tex]Na_{2}SO_{4}[/tex],

Now,

The moles of [tex]BaCl_{2}[/tex]

[tex]= 2.715[/tex] × [tex]10^-3[/tex] mol  [tex]Na_{2}SO_{4}[/tex], × 1 mol [tex]BaCl_{2}[/tex]/1 mol  [tex]Na_{2}SO_{4}[/tex],

[tex]= 2.715[/tex] × [tex]10^-3[/tex] mol  [tex]BaCl_{2}[/tex]

Mortality of the[tex]BaCl_{2}[/tex] solution:

[tex]41.25 mL[/tex] ⇒

[tex]41.25 / 1000 \\= 0.04125 L[/tex]

[tex]= 2.715[/tex] × [tex]10^-3[/tex] mol  [tex]BaCl_{2}[/tex] /[tex]0.04125 L[/tex]

[tex]= 0.0658 M[/tex]

Learn more about "Molarity" here:

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