Answer: Concentration of Cl⁻ in the final solution is 0.300 M
Step-by-step explanation:
To calculate : concentration of Cl⁻(aq) in the final solution is equal to moles of Cl⁻(aq) divide by the total volume of the solution.
i)So, number of moles of Cl⁻(aq) that are present in the HCl solution as:
moles of solute = molarity x volume (in liters)
moles of HCl = 1.00 M x 0.100 L = 0.100 moles of HCl
Since, HCl dissociates completely in water to form H⁺(aq) and Cl⁻(aq), Hence, 0.100 moles of Cl⁻(aq) present in the HCl solution.
ii) determine the number of moles of Cl⁻(aq) present in the NaCl solution.
Since, NaCl dissociates to form Na⁺(aq) and Cl⁻(aq), we have:
moles of NaCl = 2.00 M x 0.100 L = 0.200 moles of NaCl
Hence, moles of Cl⁻(aq) = 0.200 moles of NaCl
So, the total number of moles of Cl⁻(aq) in the final solution is:
moles of Cl⁻(aq) = 0.100 moles of HCl + 0.200 moles of NaCl = 0.300 moles of Cl⁻(aq)
As we know, concentration of Cl⁻(aq) = moles of Cl⁻(aq) / total volume
given Total volume = 1.00 L
Hence, concentration of Cl⁻(aq) = 0.300 moles / 1.00 L = 0.300 M
Therefore, the concentration of Cl⁻(aq) in the final solution is 0.300 M.