25.00 mL of a 0.100 M solution of HCl is added to 25.00 mL of a 0.180 M solution of NaOH - VCE - SSCE Chemistry - Question 10 - 2004 - Paper 1

Similarly, the moles of NaOH are calculated as follows:

For NaOH: $\text{moles of NaOH} = 0.180 \: \text{M} \times 0.0250 \: \text{L} = 0.00450 \: \text{moles}$

Since HCl and NaOH react in a 1:1 molar ratio, we compare the moles:

• HCl: 0.00250 moles
• NaOH: 0.00450 moles

HCl is the limiting reactant, and it will be completely consumed.

After the reaction, the remaining moles of NaOH can be calculated:

$\text{Remaining moles of NaOH} = 0.00450 - 0.00250 = 0.00200 \: \text{moles}$

The total volume of the solution after mixing is: $\text{Total Volume} = 25.00 \, \text{mL} + 25.00 \text{ mL} = 50.00 \text{ mL} = 0.0500 \text{ L}$

Now we can calculate the concentration of the remaining OH⁻ ions:

$\text{Concentration of OH}^- = \frac{\text{Remaining moles of NaOH}}{\text{Total Volume}} = \frac{0.00200 \: \text{moles}}{0.0500 \: \text{L}} = 0.0400 \: \text{M}$

The concentration of OH⁻(aq) remaining in the solution is 0.0400 M, which corresponds to option A.