Answer:
0.918 M
Step-by-step explanation:
Assuming the question requires we calculate the Molarity of sulfuric acid:
We are given:
- Volume of the acid, H₂SO₄ = 25.00 ml
- Volume of the base, KOH = 27.00 mL
- Molarity of the base, KOH is 1.70 M
We can calculate the molarity of the acid using the following steps;
Step 1: Write the chemical equation for the reaction.
The reaction is an example of a neutralization reaction where a base reacts with an acid to form salt and water.
Therefore, the balanced equation will be;
H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)
Step 2: Determine the moles of the base, KOH
When given molarity and the volume of a solution, the number of moles can be calculated by multiplying molarity with volume.
Number of moles = Molarity × Volume
= 1.700 M × 0.027 L
= 0.0459 moles
Thus, moles of KOH used is 0.0459 moles
Step 3: Determine the number of moles of the Acid, H₂SO₄
From the reaction, 1 mole of the acid reacts with 2 moles of KOH
Therefore, the mole ratio of H₂SO₄ to KOH is 1 : 2
Thus, moles of H₂SO₄ = Moles of KOH ÷ 2
= 0.0459 moles ÷ 2
= 0.02295 moles
Step 4: Calculate the molarity of the Acid
Molarity is the concentration of a solution in moles per liter
Molarity = Moles ÷ Volume
Molarity of the acid = 0.02295 moles ÷ 0.025 L
= 0.918 M
Thus, the molarity of the acid, H₂SO₄ is 0.918 M