College Chemistry Homework Help

Question

Write the balanced neutralization reaction that occurs between H₂SO₄ and KOH in aqueous solution. Phases are optional.

neutralization reaction:

Suppose 0.550 L of 0.500 M H₂SO₄ is mixed with 0.500 L of 0.200 M KOH. What concentration of sulfuric acid remains after neutralization?

concentration: 

Answer 1

We’re given two tasks:

1. Write the balanced neutralization reaction between sulfuric acid (H₂SO₄) and potassium hydroxide (KOH).

2. Calculate the concentration of sulfuric acid remaining after mixing specific volumes and concentrations of each solution.

Step 1: Understand What a Neutralization Reaction Is

A neutralization reaction occurs when an acid and a base react to form water and a salt.

- H₂SO₄ is a diprotic acid, meaning it can donate 2 protons (H⁺).

- KOH is a strong base, and each KOH molecule provides 1 hydroxide ion (OH⁻).

So to neutralize all the H⁺ from one H₂SO₄ molecule, we need 2 KOH molecules.

Balanced equation:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Tip: Always check the number of hydrogen ions your acid can donate and how many OH⁻ your base provides!

Step 2: Convert Volume to Moles

Use the formula: moles = molarity × volume

- H₂SO₄: 0.550 L × 0.500 mol/L = 0.275 mol

- KOH: 0.500 L × 0.200 mol/L = 0.100 mol

Tip: Make sure units cancel out correctly. L × mol/L gives mol!

Step 3: Determine the Limiting Reactant

From the balanced equation:

1 mol H₂SO₄ reacts with 2 mol KOH

So if we have 0.100 mol KOH, we need:

0.100 / 2 = 0.050 mol H₂SO₄

We have 0.275 mol H₂SO₄, so KOH is the limiting reactant.

Only 0.050 mol of H₂SO₄ will be used up.

Remaining H₂SO₄: 0.275 - 0.050 = 0.225 mol

Step 4: Find the Final Concentration of H₂SO₄

- Remaining H₂SO₄: 0.225 mol

- Total volume: 0.550 L + 0.500 L = 1.050 L

Concentration = 0.225 mol / 1.050 L = 0.214 M

Final Answers Recap

Balanced Reaction: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Concentration of H₂SO₄ remaining: 0.214 M