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>
To know the acidity of a
solution, we calculate the pH value. The formula for pH is given as:
<span>pH = - log [H+] where H+ must be in Molar</span>
We are given that H+ = 3.25 × 10-2 M
Therefore the pH is:
pH = - log [3.25 × 10-2]
pH = 1.488
Since pH is way below 7, therefore the solution
is acidic.
To find for the OH- concentration, we must
remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,
[H+]*[OH-] = 10^-14 <span>
</span>[OH-] = 10^-14 / [H+]
[OH-] = 10^-14 / 3.25 × 10-2
[OH-] = 3.08 × 10-13 M
Answers:
Acidic
[OH-] = 3.08 <span>× 10-13 M</span>
