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Given that it requires 37.9 mL of [tex]0.340 \, M[/tex] Na₂S₂O₃ to titrate a 20.0 mL sample of [tex]I_3^-[/tex], what is the concentration of [tex]I_3^-[/tex] in the sample?

Answer :

Final answer:

Chemical Titration The concentration of I₃- in the sample is calculated by the formula c1v1 = c2v2 and finding the missing c2. Given the information, we find that the concentration of I₃- in the sample is 0.32M.

Explanation:

The question provided involves a chemical titration between Sodium thiosulfate, Na₂S₂O₃, and iodine ion, I3-. The balance equation for this reaction can be represented as 2Na₂S₂O₃ (aq) + I₃- (aq) → S4O₆²- (aq) + 3NaI (aq). For every one mole of I₃-, two moles of Na2S2O3 are needed.

In order to find the concentration of unknown I₃-, we use the formula c1v1 = c2v2 where c1 is the concentration of Na2S2O3, v1 is the volume of Na2S2O3, c2 is the concentration of I₃- (which we aim to find), and v2 is the volume of I₃-. Hence the concentration, c2 = (c1*v1) / 2*v2. Substituting the given values we get c2 = (0.340 mol/L * 37.9 mL) / (2 * 20.0 mL) = 0.32M.

Hence, the concentration of I₃- in the sample is 0.32M.

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