Answer:
0.0827M of H₂SO₃
Explanation:
LiOH reacts with H₂SO₃ to produce water and Li₂SO₃, thus:
2LiOH + H₂SO₃ → 2H₂O + Li₂SO₃
<em>Where 2 moles of lithium hydroxide react with 1 mole of sulfurous acid.</em>
As the chemist requires 22.14mL = 0.02214L of a 0.210M solution to neutralize the acid, moles of LiOH are:
0.02214L × (0.210mol / L) =<em>0.004649 moles of LiOH</em>.
As 2 moles of LiOH react with 1 mole of H₂SO₃, moles of H₂SO₃ are:
0.004649 moles of LiOH ₓ (1 mole H₂SO₃ / 2 mol LiOH) =
<em>0.002325 moles of H₂SO₃</em>
These moles are present in 28.10mL = 0.02810L. Thus, molar concentration of the acid is:
0.002325 moles H₂SO₃ / 0.02810L = <em>0.0827M of H₂SO₃</em>