Question

3. A solution is prepared by adding 100.0 ml of 1.00 M HCl to 100.0ml of 2.00 M NaCl and

diluting to a total final volume of 1.00 L.The concentration of Cl(aq) in the final solution is,
(A) 0.100 M
(B) 0.200 M
(C) 0.300 M
(D) 2.00 M

Answer 1

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.