Titration Calculations

Part of the Acids and Bases section of Coddy's Chemistry journey. Lesson 22 of 30.

Three steps: (1) n = c × V of the known solution; (2) the ratio of the equation; (3) c = n / V or m = n × M.

25 mL of NaOH needs 20 mL of 0.1 M HCl: 0.002 mol acid, so 0.002 mol base, and c = 0.002 / 0.025 = 0.08 M.

Challenge

$25\ \mathrm{mL}$ of sodium hydroxide solution is neutralized by $20\ \mathrm{mL}$ of $0.1\ \mathrm{M}$ HCl. What is the concentration of the sodium hydroxide?

Try it yourself

Solve for $c_{b}$

Test yourself

This lesson includes a short quiz. Start the lesson to answer it and track your progress.

All lessons in Acids and Bases

1Acids and Bases

What Is an Acid What Is a Base Proton Transfer Conjugate Pairs Water as Acid and Base Recap - Proton Swap

4Neutralization

Neutralization Net Ionic Equations Salts TitrationTitration CalculationsRecap - Titration Lab

2Strong and Weak

Strong Acids Weak Acids Ions in Acid Solutions Recap - Acid Strength

5Acids Around Us

Indicators Acid Rain Antacids Recap - Soil and Lime

3The pH Scale

Ions of Water The pH Scale pH of Strong Acids pH of Strong Bases pOH Diluting an Acid Recap - pH Meter

6Practice Problems

Recap - Vinegar Test Recap - Stomach Acid Recap - Drain Cleaner