The molar concentration of the KI_3 solution is 0.0833 mol/L.
Step 1. Calculate the moles of S_2O_3^(2-)
Moles of S_2O_3^(2-) = 25.00 mL S_2O_3^(2-) ×[0.200 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)] = 5.000 mmol S_2O_3^(2-)
Step 2. Calculate the moles of I_3^(-)
Moles of I_3^(-) = 5.000 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 2.500 mmol I_3^(-)
Step 3. Calculate the molar concentration of the I_3^(-)
c = "moles"/"litres" = 2.500 mmol/30.00 mL = 0.083 33 mol/L
Answer: 0.083 M
Explanation:
[tex]2S_2O_3^{2-}(aq)+I_3^-(aq)\rightarrow S_4O_6^{2-}(aq) + 3I^-(aq)[/tex]
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{moles}{\text {Volume in L}}[/tex]
moles of [tex]K_2S_2O_3=Molarity\times {\text {Volume in L}}=0.200\times 0.025=5\times 10^{-3}moles[/tex]
According to stoichiometry: 2 moles of [tex]2S_2O_3^{2-}(aq)[/tex] require 1 mole of [tex]I_3^-[/tex]
Thus [tex]5\times 10^{-3}moles[/tex] require=[tex]\frac{1}{2}\times 5\times10^{-3}=2.5\times 10^{-3}[/tex] moles of [tex]I_3^-[/tex]
Thus Molarity of [tex]I_3^-=\frac{2.5\times 10^{-3}}{0.030L}=0.083M[/tex]
