What volume of \(0.149 \mathrm{M} \mathrm{HCl}\) must be added to \(1.00 \times 10^{2} \mathrm{mL}\) of \(0.285 \mathrm{M} \mathrm{HCl}\) so that the resulting solution has a molarity of \(0.205 \mathrm{M} ?\) Assume that the volumes are additive.

From the problem, we are given: \n- The molarity of the original solution (\(0.285 \mathrm{M}\)).\n- The volume of the original solution \(1.00 \times 10^{2} \mathrm{mL}\). \n- The molarity of the weaker solution that we're adding (\(0.149 \mathrm{M}\)). \n- The molarity of the final solution (\(0.205 \mathrm{M}\)). \nThe task is to determine the volume of the \(0.149 \mathrm{M} \mathrm{HCl}\) solution to be added to the original solution.

Considering that the molarity \((M)\) of a solution equals the moles of solute \((n)\) divided by the volume of solution \((V)\) in liters, we can write the equation for the molarity and volume before and after mixing the two solutions as: \n\n\(M_{1}V_{1} + M_{2}V_{2} = M_{f}(V_{1} + V_{2})\). \n\nWhere: \n- \(M_{1}\) and \(V_{1}\) are the molarity and volume of the original solution, \n- \(M_{2}\) and \(V_{2}\) are the molarity and volume of the weaker solution that we're adding, \n- \(M_{f}\) is the molarity of the final solution. \n\nThe volume \(V_{2}\) of \(0.149 \mathrm{M} \mathrm{HCl}\) solution to be added can be found by rearranging to: \n\n\(V_{2} = \frac{M_{f}(V_{1} + V_{2}) - M_{1}V_{1}}{M_{2}}\).

Substituting the given values into the equation: \n\n\(V_{2} = \frac{0.205 \mathrm{M} \times (1.00 \times 10^{2} \mathrm{mL} + V_{2}) - 0.285 \mathrm{M} \times 1.00 \times 10^{2} \mathrm{mL}}{0.149 \mathrm{M}}\). \n\nSolving above equation for \(V_{2}\), seek for help from tools such as a mathematical calculator to simplify the equation and calculate the volume.

The final volume \(V_{2}\) calculated in Step 3 will be in mL. Make sure to convert it to the most appropriate unit, for example, liters (if asked)