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

To find the moles of NaOH, use the same formula: For NaOH: $\text{Moles of NaOH} = 0.180 \text{ M} \times 0.02500 \text{ L} = 0.00450 \text{ moles}$

HCl and NaOH react in a 1:1 ratio: $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ Since we have 0.00250 moles of HCl, it will fully react with 0.00250 moles of NaOH. Therefore, the remaining moles of NaOH are: $\text{Remaining NaOH} = 0.00450 - 0.00250 = 0.00200 \text{ moles}$

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

To find the concentration of OH⁻ remaining in the solution, use: $\text{Concentration of OH}^- = \frac{\text{Moles of remaining NaOH}}{\text{Total Volume}}$ Substituting in the values: $\text{Concentration of OH}^- = \frac{0.00200 \text{ moles}}{0.0500 \text{ L}} = 0.0400 \text{ M}$