Equal volumes of 0.5 M HCl, 0.25 M NaOH and 2.75 M NaCl are mixed. The molarity of the $$\text{Cl}^-$$ ion and NaCl respectively

Solution

Take 1 L of each, 3 L in all. NaOH neutralises 0.25 mol of HCl, forming 0.25 mol NaCl, so NaCl totals $$2.75 + 0.25 = 3.0$$ mol, i.e. $$\frac{3.0}{3} = 1.0\text{ M}$$. Chloride from NaCl is also 1.0 M; including the 0.25 mol of HCl left over, total $$\text{Cl}^-$$ is $$\frac{3.25}{3} \approx 1.08\text{ M}$$, which is closest to option a.

Was this solution helpful?