Question

A chemist uses a standard solution of 0.210 M lithium hydroxide (LiOH) to titrate 28.10 mL of sulfurous acid acid (H2SO3), she finds that it requires 22.14 mL of the base to reach the end-point of the titration. What is the molarity of the acid solution? What is the concentration of H2SO3?
Your answer must be rounded to the correct number of significant figures. Be sure to specify a unit.

Answer 1

Answer:

$M_{acid}=0.0827M$

Explanation:

Hello,

In this case, we should consider the acid-base reaction between sulfurous acid and lithium hydroxide:

$2LiOH+H_2SO_3\rightarrow Li_2SO3+2H_2O$

Thus, we notice a 2:1 molar ratio between lithium hydroxide and sulfurous acid, for that reason, at the equivalence point we have:

$2*n_{acid}=n_{base}$

That in terms of concentrations and volumes is:

$2*M_{acid}V_{base}=M_{base}V_{base}$

Thus, we solve for the molarity of the acid which is sulfurous acid:

$M_{acid}=\frac{M_{base}V_{base}}{2*V_{base}} =\frac{0.210M*22.14mL}{2*28.10mL}\\\\M_{acid}=0.0827M$

Best regards.

Answer 2

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