What concentration of ClO3 results when 765 mL of 0.733 M AgClO2 is mixed with 789 mL of 0.337 M Mn(ClO3)2? [ClO3] 1.0514
The Correct Answer and Explanation is:
To find the concentration of ClO3−\text{ClO}_3^-ClO3− when 765 mL of 0.733 M AgClO2\text{AgClO}_2AgClO2 is mixed with 789 mL of 0.337 M Mn(ClO3)2\text{Mn(ClO}_3)_2Mn(ClO3)2, we can follow a systematic approach.
Step 1: Identify the relevant reactions
- AgClO2\text{AgClO}_2AgClO2 contains the ClO2−\text{ClO}_2^-ClO2− ion, which can react to form ClO3−\text{ClO}_3^-ClO3−.
- Mn(ClO3)2\text{Mn(ClO}_3)_2Mn(ClO3)2 contains ClO3−\text{ClO}_3^-ClO3−, which is already in its final form.
Step 2: Write the balanced reactions (if necessary)
The silver chlorite (AgClO2\text{AgClO}_2AgClO2) will undergo oxidation to form ClO3−\text{ClO}_3^-ClO3−, but for simplicity, we assume the reaction goes to completion, and the ClO2−\text{ClO}_2^-ClO2− ion is oxidized to ClO3−\text{ClO}_3^-ClO3−.
So, the primary change is the production of ClO3−\text{ClO}_3^-ClO3− from ClO2−\text{ClO}_2^-ClO2−.
Step 3: Calculate the moles of ClO2−\text{ClO}_2^-ClO2− and ClO3−\text{ClO}_3^-ClO3− from each solution
- For AgClO2\text{AgClO}_2AgClO2, the number of moles of ClO2−\text{ClO}_2^-ClO2− can be calculated as:
moles of ClO2−=0.733 M×0.765 L=0.561 mol\text{moles of } \text{ClO}_2^- = 0.733 \, \text{M} \times 0.765 \, \text{L} = 0.561 \, \text{mol}moles of ClO2−=0.733M×0.765L=0.561mol
- For Mn(ClO3)2\text{Mn(ClO}_3)_2Mn(ClO3)2, the number of moles of ClO3−\text{ClO}_3^-ClO3− is:
moles of ClO3−=0.337 M×0.789 L=0.266 mol\text{moles of } \text{ClO}_3^- = 0.337 \, \text{M} \times 0.789 \, \text{L} = 0.266 \, \text{mol}moles of ClO3−=0.337M×0.789L=0.266mol
Step 4: Add the moles of ClO3−\text{ClO}_3^-ClO3−
- From AgClO2\text{AgClO}_2AgClO2, 0.561 moles of ClO2−\text{ClO}_2^-ClO2− will be converted into ClO3−\text{ClO}_3^-ClO3− (assuming complete oxidation).
- From Mn(ClO3)2\text{Mn(ClO}_3)_2Mn(ClO3)2, 0.266 moles of ClO3−\text{ClO}_3^-ClO3− are already present.
Thus, the total moles of ClO3−\text{ClO}_3^-ClO3− is:Total moles of ClO3−=0.561 mol+0.266 mol=0.827 mol\text{Total moles of } \text{ClO}_3^- = 0.561 \, \text{mol} + 0.266 \, \text{mol} = 0.827 \, \text{mol}Total moles of ClO3−=0.561mol+0.266mol=0.827mol
Step 5: Calculate the final volume of the solution
The total volume is the sum of the individual volumes:Total volume=765 mL+789 mL=1554 mL=1.554 L\text{Total volume} = 765 \, \text{mL} + 789 \, \text{mL} = 1554 \, \text{mL} = 1.554 \, \text{L}Total volume=765mL+789mL=1554mL=1.554L
Step 6: Calculate the final concentration of ClO3−\text{ClO}_3^-ClO3−
The concentration of ClO3−\text{ClO}_3^-ClO3− is the total moles of ClO3−\text{ClO}_3^-ClO3− divided by the total volume:[ClO3−]=0.827 mol1.554 L=0.532 M[\text{ClO}_3^-] = \frac{0.827 \, \text{mol}}{1.554 \, \text{L}} = 0.532 \, \text{M}[ClO3−]=1.554L0.827mol=0.532M
However, it seems you provided an answer of 1.0514 M. To reach that, we need to carefully reconsider the reactions involved. If there was a more efficient conversion of ClO2−\text{ClO}_2^-ClO2− to ClO3−\text{ClO}_3^-ClO3− or if we assumed additional oxidation steps, the resulting concentration could match closer to 1.0514 M. However, based on the basic stoichiometry, the result is 0.532 M, which seems reasonable given the provided numbers. If more details on oxidation are available, this could be adjusted.
