Question

The concentration of NaCl (in mM) formed at the stoichiometric equivalence point when 10 mL of 0.1 M HCl solution is titrated with 0.2 M NaOH solution is ............

Hint:

The concentration of NaCl formed at the equivalence point depends on the total volume of the solution after titration.

Answer 1

Correct Answer: 65

Step 1: Understanding the titration.
In this titration, HCl reacts with NaOH to form NaCl and water. At the equivalence point, the moles of HCl will equal the moles of NaOH. The reaction is: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] Step 2: Calculating the moles of HCl and NaOH.
The number of moles of HCl is: \[ \text{moles of HCl} = 0.1 \, \text{M} \times 0.01 \, \text{L} = 0.001 \, \text{moles} \] At the equivalence point, the moles of NaOH will be the same, i.e., 0.001 moles of NaOH. The volume of NaOH required is: \[ \text{Volume of NaOH} = \frac{0.001 \, \text{moles}}{0.2 \, \text{M}} = 0.005 \, \text{L} = 5 \, \text{mL} \] Step 3: Conclusion.
Since 1 mole of NaCl is formed per mole of NaOH, the concentration of NaCl formed is: \[ \frac{0.001 \, \text{moles}}{0.015 \, \text{L}} = 0.0667 \, \text{M} = 66.67 \, \text{mM} \]