To determine the volume of 0.218 M sodium sulfate (Na₂SO₄) needed to react with exactly 25.34 mL of 0.113 M barium chloride (BaCl₂), we can follow these steps:
1. Determine the number of moles of BaCl₂:
- Given:
- Molarity of BaCl₂ ([tex]\(M_{BaCl₂}\)[/tex]) = 0.113 M
- Volume of BaCl₂ ([tex]\(V_{BaCl₂}\)[/tex]) = 25.34 mL
- Convert the volume from milliliters to liters:
[tex]\[
V_{BaCl₂} = \frac{25.34 \text{ mL}}{1000} = 0.02534 \text{ L}
\][/tex]
- Calculate the moles of BaCl₂:
[tex]\[
\text{moles of BaCl₂} = M_{BaCl₂} \times V_{BaCl₂} = 0.113 \text{ M} \times 0.02534 \text{ L} = 0.00286342 \text{ moles}
\][/tex]
2. Relate the moles of BaCl₂ to the moles of Na₂SO₄:
- From the balanced equation:
[tex]\[
\text{BaCl}_2( \text{aq} ) + \text{Na}_2\text{SO}_4( \text{aq} ) \rightarrow \text{BaSO}_4( \text{s} ) + 2 \text{NaCl}( \text{aq} )
\][/tex]
It shows a 1:1 molar ratio between BaCl₂ and Na₂SO₄.
- Therefore, the moles of Na₂SO₄ needed are equal to the moles of BaCl₂:
[tex]\[
\text{moles of Na₂SO₄} = 0.00286342 \text{ moles}
\][/tex]
3. Calculate the volume of 0.218 M Na₂SO₄ needed:
- Given:
- Molarity of Na₂SO₄ ([tex]\(M_{Na₂SO₄}\)[/tex]) = 0.218 M
- Using the formula [tex]\( \text{Molarity} = \frac{\text{moles}}{\text{volume}} \)[/tex]:
[tex]\[
\text{Volume of Na₂SO₄ (in L)} = \frac{\text{moles of Na₂SO₄}}{M_{Na₂SO₄}} = \frac{0.00286342 \text{ moles}}{0.218 \text{ M}} = 0.013134954128440368 \text{ L}
\][/tex]
- Convert the volume from liters to milliliters:
[tex]\[
\text{Volume of Na₂SO₄ (in mL)} = 0.013134954128440368 \text{ L} \times 1000 \text{ mL/L} = 13.134954128440368 \text{ mL}
\][/tex]
Hence, you need approximately 13.135 mL of 0.218 M sodium sulfate to react with 25.34 mL of 0.113 M barium chloride.
