Answer:
0.04 M.
Step-by-step explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Ba(OH)₂ + H₂SO₄ —> BaSO₄ + 2H₂O
From the balanced equation above,
The mole ratio of acid, H₂SO₄ (nₐ) = 1
The mole ratio of base, Ba(OH)₂ (n₆) = 1
Finally, we shall determine the molarity of Ba(OH)₂ solution. This can be obtained as follow:
Volume of base, Ba(OH)₂ (V₆) = 93.9 mL
Volume of acid, H₂SO₄ (Vₐ) = 15.3 mL
Molarity of acid, H₂SO₄ (Mₐ) = 0.247 M
Molarity of base, Ba(OH)₂ (M₆) =?
MₐVₐ / M₆V₆ = nₐ / n₆
0.247 × 15.3 / M₆ × 93.9 = 1/1
3.7791 / M₆ × 93.9 = 1
Cross multiply
M₆ × 93.9 = 3.7791
Divide both side by 93.9
M₆ = 3.7791 / 93.9
M₆ = 0.04 M
Therefore, the molarity of the Ba(OH)₂ solution is 0.04 M