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Answer :
The final concentration of both Na⁺ and OH⁻ ions in the mixed solution of NaOH is 0.278 M.
To calculate the final concentration of Na⁺ and OH⁻ ions, we must use the formula M₁V₁ = M₂V₂, where M₁ and V₁ are the molarity and volume of the first solution and M₂ and V₂ are the molarity and volume of the second solution respectively. This is a dilution calculation because we are mixing two volumes of the same substance (NaOH).
First, we convert the volumes to liters (L) and then we calculate the total number of moles of NaOH added:
- For the 0.170 M NaOH: (0.170 moles/L) × (0.0420 L) = 0.00714 moles.
- For the 0.400 M NaOH: (0.400 moles/L) × (0.0376 L) = 0.01504 moles.
Now, we sum up the moles of NaOH and divide by the total volume to find the molarity of the final solution:
- Total moles of NaOH = 0.00714 moles + 0.01504 moles = 0.02218 moles.
- Total volume in liters = 42.0 mL + 37.6 mL = 79.6 mL = 0.0796 L.
- Final concentration (Molarity) = Total moles / Total Volume = 0.02218 moles / 0.0796 L = 0.278 M NaOH.
Since NaOH dissociates completely in aqueous solution into Na⁺ and OH⁻ ions, the final concentration of each ion (Na⁺ and OH⁻) in the solution is also 0.278 M.
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Rewritten by : Barada
The final concentration of both Na+ and OH- ions will be 0.2787 M after mixing 42.0 mL of 0.170 M NaOH with 37.6 mL of 0.400 M NaOH, calculated by finding the total moles of NaOH added and dividing by the total volume of the mixture in liters.
The question involves mixing two solutions of NaOH with different concentrations and calculating the concentration of ions in the resulting solution. To calculate the final concentration of Na+ and OH- ions, we first need to find the total number of moles of NaOH added and then divide by the total volume of the solution in liters.
Calculate the moles of NaOH in each solution before mixing:
Moles from 42.0 mL of 0.170 M NaOH = 0.170 mol/L × 0.0420 L = 0.00714 mol
Moles from 37.6 mL of 0.400 M NaOH = 0.400 mol/L × 0.0376 L = 0.01504 mol
Add these to find the total moles of NaOH:
Total moles of NaOH = 0.00714 mol + 0.01504 mol = 0.02218 mol
Calculate the total volume of the mixture in liters:
Total volume = 42.0 mL + 37.6 mL = 79.6 mL = 0.0796 L
Now we find the final concentration by dividing the total moles by the total volume:
Final concentration of NaOH = 0.02218 mol / 0.0796 L = 0.2787 M
Since NaOH dissociates completely into Na+ and OH- ions in solution, the concentration of Na+ and OH- will both be 0.2787 M after mixing.
