What is the concentration of the solution that results from mixing `40.0` mL of `0.200` M HCl with `60.0` mL of `0.100` M NaOH?(You may assume the volume are additive.)

Options

A. `0.150` M NaCl
B. `0.0200` M NaCl and `0.0200` M HCl
C. `0.0200` M NaCl and `0.0600` M HCl
D. 0.0600 N NaCl and 0.0200 M HCl

Answer

To find the concentration of the solution resulting from mixing `40.0` mL of `0.200` M HCl with `60.0` mL of `0.100` M NaOH, we can follow these steps:

### Step 1: Calculate the moles of HCl and NaOH

1. **Calculate moles of HCl:**
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.200 \, \text{mol/L} \times 0.0400 \, \text{L} = 0.00800 \, \text{mol} \]

2. **Calculate moles of NaOH:**
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.100 \, \text{mol/L} \times 0.0600 \, \text{L} = 0.00600 \, \text{mol} \]

### Step 2: Determine the limiting reactant

- The balanced chemical equation for the reaction is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

- From the equation, we see that HCl and NaOH react in a 1:1 ratio.

- We have:
- Moles of HCl = 0.00800 mol
- Moles of NaOH = 0.00600 mol

- Since NaOH is the limiting reactant (less moles), it will determine the amount of product formed.

### Step 3: Calculate the moles of excess HCl

- After the reaction, we will have some HCl left over:
\[ \text{Excess HCl} = \text{Initial moles of HCl} - \text{Moles of NaOH} = 0.00800 \, \text{mol} - 0.00600 \, \text{mol} = 0.00200 \, \text{mol} \]

### Step 4: Calculate the total volume of the solution

- Total volume after mixing:
\[ \text{Total Volume} = 40.0 \, \text{mL} + 60.0 \, \text{mL} = 100.0 \, \text{mL} = 0.100 \, \text{L} \]

### Step 5: Calculate the concentration of the remaining HCl

- The concentration of the remaining HCl in the solution:
\[ \text{Concentration of HCl} = \frac{\text{Moles of excess HCl}}{\text{Total Volume}} = \frac{0.00200 \, \text{mol}}{0.100 \, \text{L}} = 0.0200 \, \text{M} \]

### Final Answer

The concentration of the solution after mixing `40.0` mL of `0.200` M HCl with `60.0` mL of `0.100` M NaOH is **0.0200 M**.

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