Question
Liquid-Liquid Titration problem with 30.00 mL of 0.150 M NaOH and 20.00 mL of unknown concentration HCl acid solution
Answers
GPT-4o mini
To solve a liquid-liquid titration problem involving NaOH and HCl, we first need to set up the relevant chemical reaction and then calculate the unknown concentration of the HCl solution.
The neutralization reaction between NaOH (a strong base) and HCl (a strong acid) can be represented by the following balanced equation:
\[
\text{NaOH}_{(aq)} + \text{HCl}_{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)}
\]
From this equation, we see that the stoichiometry of the reaction is 1:1, meaning one mole of NaOH reacts with one mole of HCl.
### Step 1: Calculate Moles of NaOH Used
Given that you have 30.00 mL of 0.150 M NaOH, we can calculate the moles of NaOH present:
1. Convert mL to L:
\[
30.00 \text{ mL} = 0.03000 \text{ L}
\]
2. Calculate moles of NaOH:
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.03000 \text{ L} \times 0.150 \text{ mol/L} = 0.00450 \text{ mol}
\]
### Step 2: Set Up the Stoichiometric Relationship
Since the stoichiometry of the reaction is 1:1, the moles of HCl will also be:
\[
\text{Moles of HCl} = \text{Moles of NaOH} = 0.00450 \text{ mol}
\]
### Step 3: Calculate the Concentration of HCl
You have 20.00 mL of HCl solution. To find the concentration of HCl, we will convert the volume from mL to L and divide the moles of HCl by this volume:
1. Convert mL to L:
\[
20.00 \text{ mL} = 0.02000 \text{ L}
\]
2. Calculate the concentration of HCl:
\[
\text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl (L)}} = \frac{0.00450 \text{ mol}}{0.02000 \text{ L}} = 0.225 \text{ M}
\]
### Final Answer
The concentration of the unknown HCl solution is **0.225 M**.
The neutralization reaction between NaOH (a strong base) and HCl (a strong acid) can be represented by the following balanced equation:
\[
\text{NaOH}_{(aq)} + \text{HCl}_{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)}
\]
From this equation, we see that the stoichiometry of the reaction is 1:1, meaning one mole of NaOH reacts with one mole of HCl.
### Step 1: Calculate Moles of NaOH Used
Given that you have 30.00 mL of 0.150 M NaOH, we can calculate the moles of NaOH present:
1. Convert mL to L:
\[
30.00 \text{ mL} = 0.03000 \text{ L}
\]
2. Calculate moles of NaOH:
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.03000 \text{ L} \times 0.150 \text{ mol/L} = 0.00450 \text{ mol}
\]
### Step 2: Set Up the Stoichiometric Relationship
Since the stoichiometry of the reaction is 1:1, the moles of HCl will also be:
\[
\text{Moles of HCl} = \text{Moles of NaOH} = 0.00450 \text{ mol}
\]
### Step 3: Calculate the Concentration of HCl
You have 20.00 mL of HCl solution. To find the concentration of HCl, we will convert the volume from mL to L and divide the moles of HCl by this volume:
1. Convert mL to L:
\[
20.00 \text{ mL} = 0.02000 \text{ L}
\]
2. Calculate the concentration of HCl:
\[
\text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl (L)}} = \frac{0.00450 \text{ mol}}{0.02000 \text{ L}} = 0.225 \text{ M}
\]
### Final Answer
The concentration of the unknown HCl solution is **0.225 M**.