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What is the concentration of chloride ions, in mol dm−3, in a solution formed by mixing 200 cm3of 1 mol dm−3 HCl with 200 cm3 of 5 mol dm−3 NaCl?

Question

What is the concentration of chloride ions, in mol dm−3, in a solution formed by mixing 200 cm³ of 1 mol dm−3 HCl with 200 cm³ of 5 mol dm−3 NaCl?

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Solution

To find the concentration of chloride ions in the solution, we first need to find the total amount of chloride ions contributed by both HCl and NaCl.

  1. Calculate the moles of chloride in each solution:

    For HCl: Moles = concentration x volume = 1 mol/dm^3 x 0.2 dm^3 = 0.2 mol For NaCl: Moles = concentration x volume = 5 mol/dm^3 x 0.2 dm^3 = 1 mol

  2. Add the moles of chloride from both solutions to get the total moles of chloride:

    Total moles of chloride = moles from HCl + moles from NaCl = 0.2 mol + 1 mol = 1.2 mol

  3. Calculate the total volume of the solution:

    Total volume = volume of HCl + volume of NaCl = 0.2 dm^3 + 0.2 dm^3 = 0.4 dm^3

  4. Finally, calculate the concentration of chloride ions in the solution:

    Concentration = total moles / total volume = 1.2 mol / 0.4 dm^3 = 3 mol/dm^3

So, the concentration of chloride ions in the solution is 3 mol/dm^3.

This problem has been solved

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