Question:
What is the molarity of a potassium triiodide solution, {eq}KI_3(aq) {/eq}, if {eq}30.00 mL {/eq} of the solution is required to completely react with {eq}25.00 mL {/eq} of a {eq}0.200 mol L^{-1} {/eq} thiosulfate solution, {eq}K_2S_2O_3(aq) {/eq}? The chemical equation for the reaction is {eq}2S_2O_3^{2-}(aq)+ I_3^-(aq)to S_4O_6^{2-}(aq)+ 3I^-(aq) {/eq}
A)0.0833 {eq}mol L^{-1} {/eq}
B)0.120 {eq}mol L^{-1} {/eq}
C)0.167 {eq}mol L^{-1} {/eq}
D)0.333 {eq}mol L^{-1} {/eq}
The balanced chemical equation for the reaction is:
2S2O3^2-(aq) + I3^-(aq) ? S4O6^2-(aq) + 3I^-(aq)
From the equation, we can see that 1 mole of I3^- reacts with 2 moles of S2O3^2-. Therefore, the number of moles of S2O3^2- in 25.00 mL of 0.200 mol L^-1 solution is:
n(S2O3^2-) = C(V) = (0.200 mol L^-1)(0.02500 L) = 0.00500 mol
Since 1 mole of I3^- reacts with 2 moles of S2O3^2-, the number of moles of I3^- in 25.00 mL of 0.200 mol L^-1 solution is:
n(I3^-) = 0.5n(S2O3^2-) = 0.5(0.00500 mol) = 0.00250 mol
Now we need to find the molarity of the KI3 solution that reacts with 0.00250 mol of I3^- in 30.00 mL of the solution. The molarity is given by:
Molarity = n/V
where n is the number of moles of solute and V is the volume of solution in liters. Therefore, the molarity of the KI3 solution is:
Molarity = n/V = (0.00250 mol)/(0.03000 L) = 0.0833 mol L^-1
Therefore, the answer is A) 0.0833 mol L^-1.