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–?

1.87x10⁻³ M SO₄²⁻

Explanation:

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

SO₄²⁻(aq) + Ba²⁺(aq) → BaSO₄(s)

Where 1 mole of  SO₄²⁻ reacts per mole of Ba²⁺

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).

Moles Ba²⁺:

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

Molarity SO₄²⁻:

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

1.87x10⁻⁴ moles of SO₄²⁻ / 0.1L =

A base

Explanation:

NaOH is an Arrhenius base because they yield the hydroxide ion (OH-) when they ionize.