Refer To The Precipitation Reaction Below. BaCl2 (Aq) + H2SO4 (Aq) – BaSO4 (S) + 2HCl (Aq) How Much 0.22 M H2SO4 Solution In Liters Will Completely Precipitate The Ba²? Ion In 1.2 L Of 0.19 M BaCl2 Solution? Do Not Include Units In Your Answer, And Round To Two Significant Figures.
Refer to the precipitation reaction below. BaCl2 (aq) + H2SO4 (aq) – BaSO4 (s) + 2HCl (aq) How much 0.22 M H2SO4 solution in
The Correct Answer and Explanation is:
Answer: 1.0
Step-by-Step Solution and Explanation
We are given the balanced chemical reaction:
BaCl₂ (aq) + H₂SO₄ (aq) → BaSO₄ (s) + 2HCl (aq)
This shows a 1:1 molar ratio between BaCl₂ and H₂SO₄, meaning one mole of sulfuric acid reacts with one mole of barium chloride to form one mole of barium sulfate precipitate.
Step 1: Calculate moles of Ba²⁺ in BaCl₂ solution
We are told:
- Volume of BaCl₂ = 1.2 L
- Molarity of BaCl₂ = 0.19 M
Moles of BaCl2=Molarity×Volume=0.19 mol/L×1.2 L=0.228 mol\text{Moles of BaCl}_2 = Molarity \times Volume = 0.19\ \text{mol/L} \times 1.2\ \text{L} = 0.228\ \text{mol}
Since each BaCl₂ provides one Ba²⁺ ion, moles of Ba²⁺ = 0.228 mol.
Step 2: Use stoichiometry to find moles of H₂SO₄ needed
From the balanced equation:
- 1 mole of H₂SO₄ reacts with 1 mole of Ba²⁺.
- So, moles of H₂SO₄ needed = 0.228 mol
Step 3: Use molarity of H₂SO₄ to find required volume
We are told:
- Molarity of H₂SO₄ = 0.22 M
Volume of H2SO4=MolesMolarity=0.228 mol0.22 mol/L=1.036 L\text{Volume of H}_2\text{SO}_4 = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.228\ \text{mol}}{0.22\ \text{mol/L}} = 1.036\ \text{L}
Step 4: Round the final answer
Rounded to two significant figures, the answer is: 1.0\boxed{1.0}
Final Explanation:
To completely precipitate all the Ba²⁺ ions from a 1.2 L solution of 0.19 M BaCl₂, you need the same number of moles of H₂SO₄ because the reaction has a 1:1 molar ratio. You calculated the moles of Ba²⁺ present, matched it to the moles of H₂SO₄ needed, and then used the molarity of H₂SO₄ to find the required volume. Rounding gives a final answer of 1.0 liters.
