To determine whether the learner's solution has the correct concentration after adding 200 cm³ of pure water to the dissolved sodium hydroxide (NaOH) pellets, we need to consider the definitions of concentration and how dilution affects it.
Given:
- Required volume of NaOH solution: 200 cm³
- Desired concentration of NaOH solution: 0.5 mol/dm³ (which is the same as 0.5 M)
Calculating the required moles of NaOH:
The first step is to find out how many moles of NaOH are needed in 200 cm³ to achieve a concentration of 0.5 M.
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Convert the volume from cm³ to dm³: \[ 200 \text{ cm}^3 = 0.200 \text{ dm}^3 \]
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Calculate the required moles using the desired concentration: \[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume (in dm}^3) \] \[ \text{Moles of NaOH} = 0.5 , \text{mol/dm}^3 \times 0.200 , \text{dm}^3 = 0.1 , \text{mol} \]
Calculating the mass of NaOH required:
Now, calculate the mass needed to provide this number of moles. The molar mass of NaOH is approximately 40 g/mol.
- Calculate the required mass: \[ \text{Mass of NaOH} = \text{Moles} \times \text{Molar Mass} = 0.1 , \text{mol} \times 40 , \text{g/mol} = 4 , \text{g} \]
Mixing with pure water:
The learner accurately measured the 4 g NaOH and dissolved it in a volume of pure water to make a total of 200 cm³ of solution.
However, when the learner adds 200 cm³ of pure water after dissolving the NaOH, the total volume of the solution changes:
- New total volume after adding water: \[ \text{Total Volume} = 200 , \text{cm}^3 \text{ (initial solution)} + 200 , \text{cm}^3 \text{ (added water)} = 400 , \text{cm}^3 \]
Calculating the new concentration after dilution:
The concentration of the solution can be calculated from the total number of moles of NaOH and the new total volume.
- Concentration after dilution: \[ \text{New Concentration} = \frac{\text{Moles of NaOH}}{\text{Total Volume (in dm}^3)} \] The total volume in dm³ is: \[ 400 , \text{cm}^3 = 0.400 , \text{dm}^3 \] Using the moles calculated earlier (0.1 mol): \[ \text{New Concentration} = \frac{0.1 , \text{mol}}{0.400 , \text{dm}^3} = 0.25 , \text{mol/dm}^3 \]
Conclusion:
After adding the 200 cm³ of pure water, the final concentration of the NaOH solution will be 0.25 M, which is lower than the desired concentration of 0.5 M. Therefore, the solution will not have the correct concentration due to the dilution that occurred by adding additional water.
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