Question

(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) Suppose 0.750 L of 0.480 M H2SO4 is mixed with 0.700 L of 0.290 M KOH. What concentration of sulfuric acid remains after neutralization?

Answer 1

These are two questions and two answers

Answer:

    Question 1:

• H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)

    Question 2:

• 0.201 M

Explanation:

Question 1:

The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) Word equation:

• sulfuric acid + potassium hydroxide → potassium sulfate + water

                 ↑                               ↑                              ↑                       ↑

               acid                          base                        salt                   water

2) Skeleton equation (unbalanced)

• H₂SO₄ + KOH → K₂SO₄ + H₂O

#) Balanced chemical equation (including phases)

• H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

Question 2:

1) Mol ratio:

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

• 1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

2) Moles of H₂SO₄:

• V = 0.750 liter
• M = 0.480 mol/liter
• M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

3) Moles of KOH:

• V = 0.700 liter
• M = 0.290 mol/liter
• M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

4) Determine the limiting reagent:

a) Stoichiometric ratio:

   1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

   0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄  is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

5) Amount of H₂SO₄ that reacts:

• Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

         x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

         x = 0.203 / 2 = 0.0677 mol of H₂SO₄

6) Concentration of H₂SO₄ remaining:

• Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

• Total volume = 0.700 liter + 0.750 liter = 1.450 liter

• Concetration = M

        M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer