Answer:
0.0827M of H₂SO₃
Explanation:
LiOH reacts with H₂SO₃ to produce water and Li₂SO₃, thus:
2LiOH + H₂SO₃ → 2H₂O + Li₂SO₃
Where 2 moles of lithium hydroxide react with 1 mole of sulfurous acid.
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) =0.004649 moles of LiOH.
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) =
0.002325 moles of H₂SO₃
These moles are present in 28.10mL = 0.02810L. Thus, molar concentration of the acid is:
0.002325 moles H₂SO₃ / 0.02810L = 0.0827M of H₂SO₃
Answer:
[tex]M_{acid}=0.0827M[/tex]
Explanation:
Hello,
In this case, we should consider the acid-base reaction between sulfurous acid and lithium hydroxide:
[tex]2LiOH+H_2SO_3\rightarrow Li_2SO3+2H_2O[/tex]
Thus, we notice a 2:1 molar ratio between lithium hydroxide and sulfurous acid, for that reason, at the equivalence point we have:
[tex]2*n_{acid}=n_{base}[/tex]
That in terms of concentrations and volumes is:
[tex]2*M_{acid}V_{base}=M_{base}V_{base}[/tex]
Thus, we solve for the molarity of the acid which is sulfurous acid:
[tex]M_{acid}=\frac{M_{base}V_{base}}{2*V_{base}} =\frac{0.210M*22.14mL}{2*28.10mL}\\\\M_{acid}=0.0827M[/tex]
Best regards.
