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kolezko [41]
3 years ago
8

To determine the concentration of SO4 2– ion in a sample of groundwater, 100.0 mL of the sample is titrated with 0.0250 M Ba(NO3

)2, forming insoluble BaSO4. If 7.48 mL of the Ba(NO3)2 solution is required to reach the end point of the titration, what is the molarity of the SO4 2–?
Chemistry
1 answer:
Brrunno [24]3 years ago
8 0

Answer:

1.87x10⁻³ M SO₄²⁻

Explanation:

The reaction of SO₄²⁻ with Ba²⁺ (From Ba(NO₃)₂) is:

SO₄²⁻(aq) + Ba²⁺(aq) → BaSO₄(s)

<em>Where 1 mole of  SO₄²⁻ reacts per mole of Ba²⁺</em>

<em />

To reach the end point in this titration, we need to add the same moles of Ba²⁺ that the moles that are of SO₄²⁻.

Thus, to find molarity of SO₄²⁻ we need to find first the moles of Ba²⁺ added (That will be the same of SO₄²⁻). And as the volume of the initial sample was 100mL we can find molarity (As ratio of moles of SO₄²⁻ per liter of solution).

<em>Moles Ba²⁺:</em>

7.48mL = 7.48x10⁻³L ₓ (0.0250moles / L) = 1.87x10⁻⁴ moles of Ba²⁺ = Moles of SO₄²⁻

<em>Molarity SO₄²⁻:</em>

As there are 1.87x10⁻⁴ moles of SO₄²⁻ in 100mL = 0.1L, molarity is:

1.87x10⁻⁴ moles of SO₄²⁻ / 0.1L =

<h3> 1.87x10⁻³ M SO₄²⁻</h3>
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