Titration: HCl and NaOH Calculation

What is the balanced equation for the reaction between hydrochloric acid and sodium hydroxide?

How many moles of sodium hydroxide are present in 25.0 cm³ of 0.1 mol/dm³ NaOH?

Given that the reaction between HCl and NaOH is a 1:1 reaction, how many moles of HCl would react with 0.0025 mol of NaOH?

Calculate the concentration of the HCl solution when 0.0025 mol of HCl is present in 20.0 cm³ of solution.

What is the volume of 20.0 cm³ in dm³, and why is this conversion necessary?

Titration Question: Hydrochloric Acid & Sodium Hydroxide

Problem: You are titrating hydrochloric acid (HCl) with sodium hydroxide (NaOH).
Given Data:
- 25.0 cm³ of 0.1 mol/dm³ NaOH solution.
- 20.0 cm³ of HCl solution.
Aim: Determine the concentration of the HCl solution.

Step 1: Balanced Chemical Equation

Reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Stoichiometry: 1:1 ratio (one mole of HCl reacts with one mole of NaOH).

Step 2: Formula for Moles

Formula: Moles = Concentration (mol/dm³) × Volume (dm³)

Step 3: Calculate Moles of NaOH

Conversion: Volume in cm³ to dm³ (divide by 1000).
NaOH Data:
- Concentration = 0.1 mol/dm³.
- Volume = 25.0 cm³ = 0.025 dm³.
Moles of NaOH: 0.1 mol/dm³ × 0.025 dm³ = 0.0025 mol.

Step 4: Moles of HCl

Ratio: 1:1 reaction, so moles of HCl = moles of NaOH = 0.0025 mol.

Step 5: Calculate HCl Concentration

Formula (Rearranged): Concentration = Moles ÷ Volume (dm³)
HCl Data:
- Moles = 0.0025 mol.
- Volume = 20.0 cm³ = 0.020 dm³.
Concentration of HCl: 0.0025 mol ÷ 0.020 dm³ = 0.125 mol/dm³.

Final Answer

The concentration of the HCl solution is 0.125 mol/dm³.

This quiz focuses on the titration process involving hydrochloric acid and sodium hydroxide. You will calculate the concentration of HCl based on given volumes and concentrations. Understanding the balanced chemical equation and stoichiometry is crucial for successful problem-solving.