Question

What is the molarity of a potassium triiodide solution, ki3(aq), if 30.00 ml of the solution is required to completely react with 25.00 ml of a 0.200 m thiosulfate solution, k2s2o3(aq)? the chemical equation for the reaction is 2 s2o32-(aq) + i3-(aq) → s4o62-(aq) + 3 i-(aq). what is the molarity of a potassium triiodide solution, ki3(aq), if 30.00 ml of the solution is required to completely react with 25.00 ml of a 0.200 m thiosulfate solution, k2s2o3(aq)? the chemical equation for the reaction is 2 s2o32-(aq) + i3-(aq) → s4o62-(aq) + 3 i-(aq). 0.167 m 0.333 m 0.120 m 0.0833 m?

Answer 1

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 2

Answer: 0.083 M

Explanation:

$2S_2O_3^{2-}(aq)+I_3^-(aq)\rightarrow S_4O_6^{2-}(aq) + 3I^-(aq)$

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{moles}{\text {Volume in L}}$

moles of $K_2S_2O_3=Molarity\times {\text {Volume in L}}=0.200\times 0.025=5\times 10^{-3}moles$

According to stoichiometry: 2 moles of $2S_2O_3^{2-}(aq)$ require 1 mole of $I_3^-$

Thus $5\times 10^{-3}moles$ require=$\frac{1}{2}\times 5\times10^{-3}=2.5\times 10^{-3}$ moles of $I_3^-$

Thus Molarity of $I_3^-=\frac{2.5\times 10^{-3}}{0.030L}=0.083M$