Certainly! Let's solve the problem step-by-step:
### Problem Statement:
We are given:
- 1 ml of 1 M HCl
- 10 ml of 1 M NaOH
We need to determine:
1. The amount of salt (NaCl) formed.
2. The amount of HCl left.
3. The amount of NaOH left.
### Step-by-Step Solution:
1. Convert the volumes from milliliters to liters:
For HCl:
[tex]\[
\text{Volume of HCl in Liters} = \frac{1 \text{ ml}}{1000} = 0.001 \text{ L}
\][/tex]
For NaOH:
[tex]\[
\text{Volume of NaOH in Liters} = \frac{10 \text{ ml}}{1000} = 0.01 \text{ L}
\][/tex]
2. Calculate the moles of HCl and NaOH:
Moles of HCl (using Molarity [tex]\(M = \frac{\text{moles}}{\text{L}}\)[/tex]):
[tex]\[
\text{Moles of HCl} = 1 \text{ M} \times 0.001 \text{ L} = 0.001 \text{ moles}
\][/tex]
Moles of NaOH:
[tex]\[
\text{Moles of NaOH} = 1 \text{ M} \times 0.01 \text{ L} = 0.01 \text{ moles}
\][/tex]
3. Determine the limiting reactant and the amount of product formed:
The neutralization reaction between HCl and NaOH is:
[tex]\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\][/tex]
From the stoichiometry of the reaction, 1 mole of HCl neutralizes 1 mole of NaOH to form 1 mole of NaCl.
We compare the moles of HCl and NaOH:
- Moles of HCl = 0.001 moles
- Moles of NaOH = 0.01 moles
Since 0.001 moles of HCl are less than 0.01 moles of NaOH, HCl is the limiting reactant.
4. Calculate the amount of salt (NaCl) formed:
Since HCl is the limiting reactant, the moles of NaCl formed will be equal to the moles of HCl:
[tex]\[
\text{Moles of NaCl} = 0.001 \text{ moles}
\][/tex]
5. Determine the remaining moles of reactants:
- HCl is the limiting reactant, and all of it reacts, so the remaining moles of HCl = 0.
- Since all 0.001 moles of HCl react, the moles of NaOH that react will also be 0.001 moles. Thus, the remaining moles of NaOH = 0.01 moles - 0.001 moles = 0.009 moles.
### Final Result:
- Moles of NaCl formed = 0.001 moles
- Remaining moles of HCl = 0 moles
- Remaining moles of NaOH = 0.009 moles
Therefore, when 1 ml of 1 M HCl is mixed with 10 ml of 1 M NaOH:
- The amount of salt (NaCl) formed is 0.001 moles.
- The amount of HCl left is 0 moles.
- The amount of NaOH left is 0.009 moles.
