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?
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.
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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
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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
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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
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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.
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