Respuesta :
Answer: The concentration of iodide ion in the resulting solution is 0.225 M
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .......(1)
- For KI:
Molarity of solution = 0.30 M
Volume of solution = 6.00 mL
Putting values in equation 1, we get:
[tex]0.30M=\frac{\text{Moles of KI}\times 1000}{6.0}\\\\\text{Moles of KI}=\frac{0.30\times 6.0}{1000}=0.0018moles[/tex]
1 mole of KI produces 1 mole of [tex]K^{+}[/tex] ions and 1 mole of [tex]I^-[/tex] ions
Moles of iodide ions in KI solution = (1 × 0.0018) = 0.0018 moles
- For NaI:
Molarity of solution = 0.45 M
Volume of solution = 2.00 mL
Putting values in equation 1, we get:
[tex]0.45M=\frac{\text{Moles of NaI}\times 1000}{2.0}\\\\\text{Moles of NaI}=\frac{0.45\times 2.0}{1000}=0.0009moles[/tex]
1 mole of NaI produces 1 mole of [tex]Na^{+}[/tex] ions and 1 mole of [tex]I^-[/tex] ions
Moles of iodide ions in NaI solution = (1 × 0.0009) = 0.0009 moles
Now, calculating the chloride ions in the solution by using equation 1, we get:
Total moles of iodide ions = [0.0018 + 0.0009] = 0.0027 moles
Total volume of base solution = [6 + 4 + 2] = 12 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of }I^-\text{ ions}=\frac{0.0027mol\times 1000}{12}\\\\\text{Molarity of }I^-\text{ ions}=0.225M[/tex]
Hence, the concentration of iodide ion in the resulting solution is 0.225 M
The concentration of iodide ion in the resulting solution is 0.225 M
Molarity:
To calculate the molarity of solution, we use the equation:
Molarity = Moles of solute / Volume of solution (in L) .......(1)
For KI:
Molarity of solution = 0.30 M
Volume of solution = 6.00 mL
Putting values in equation 1, we get:
0.30 M = Mole of KI * 1000 / 6.0
Mole of KI=0.0018 moles
1 mole of KI produces 1 mole of [tex]K^+[/tex] ions and 1 mole of [tex]I^-[/tex] ions
Moles of iodide ions in KI solution = (1 * 0.0018) = 0.0018 moles
For NaI:
Molarity of solution = 0.45 M
Volume of solution = 2.00 mL
Putting values in equation 1, we get:
0.45 M = Mole of NaI * 1000 / 2.0
Mole of NaI=0.0009 moles
1 mole of NaI produces 1 mole of [tex]Na^+[/tex] ions and 1 mole of [tex]I^-[/tex] ions
Moles of iodide ions in NaI solution = (1 × 0.0009) = 0.0009 moles
Calculation of chloride ions in the solution by using equation 1, we get:
Total moles of iodide ions = [0.0018 + 0.0009] = 0.0027 moles
Total volume of base solution = [6 + 4 + 2] = 12 mL
Putting values in equation 1, we get:
Molarity of [tex]I^-[/tex] ions = 0.0027 * 1000 / 12
Molarity of [tex]I^-[/tex] ions = 0.225 M
Hence, the concentration of iodide ion in the resulting solution is 0.225 M.
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