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Chapter 4: Problem 70

Iodine reacts with thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), to give iodide ion and the tetrathionate ion, \(\mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}\). (a) Write a balanced net ionic equation for the reaction. (b) If \(25.0 \mathrm{~g}\) of iodine is dissolved in enough water to make \(1.50 \mathrm{~L}\) of solution, what volume of \(0.244 M\) sodium thiosulfate will be needed for complete reaction?

Short Answer

Expert verified
Answer: The balanced net ionic equation for the reaction is: 2 I⁻(aq) + S₄O₆²⁻(aq) → I₂(s) + 2 S₂O₃²⁻(aq). To react completely with 25.0 g of iodine, 0.807 L of a 0.244 M sodium thiosulfate solution is required.

Step by step solution

01

Write the balanced net ionic equation

The reaction between iodine and thiosulfate ion is as follows: \(\displaystyle \textrm{I}_{2}( s) +\textrm{ S}_{2}\textrm{O}_{3}^{2-}( aq) \rightarrow \textrm{I}^{-}( aq) +\textrm{ S}_{4}\textrm{O}_{6}^{2-}( aq)\) Now we need to balance the equation. We see that there are 2 iodine atoms and 2 sulfur atoms on each side, but we have 3 oxygen atoms on the left side and 6 oxygen atoms on the right side. To balance the oxygens, we can use 2 thiosulfate ions on the left side. Now the balanced net ionic equation is: \(\displaystyle 2 \textrm{I}^{-}( aq) +\textrm{ S}_{4}\textrm{O}_{6}^{2-}( aq) \rightarrow \textrm{I}_{2}( s) + 2\textrm{ S}_{2}\textrm{O}_{3}^{2-}( aq)\)
02

Calculate moles of iodine

We are given 25.0 grams of iodine and we want to calculate the moles of iodine. To do this, we will use iodine's molar mass, which is 253.8 g/mol. Moles of iodine = \(\frac{\textrm{Mass of iodine}}{\textrm{Molar mass of iodine}} = \frac{25.0 \textrm{g}}{253.8 \textrm{g/mol}} = 0.0985 \textrm{mol}\)
03

Use stoichiometry to find moles of sodium thiosulfate

According to the balanced net ionic equation, 1 mole of iodine will react with 2 moles of thiosulfate ions. So, we must find out how many moles of thiosulfate ions are required to react with 0.0985 moles of iodine. Moles of sodium thiosulfate = moles of iodine \(\times\) \(\frac{\textrm{Moles of sodium thiosulfate}}{\textrm{Moles of iodine}}\) = \(0.0985 \textrm{mol} \times \frac{2 \textrm{mol}}{1 \textrm{mol}} = 0.197 \textrm{mol}\)
04

Calculate the volume of sodium thiosulfate solution needed

We are given the molarity (M) of the sodium thiosulfate solution, which is 0.244 M. We can use this information along with the moles of sodium thiosulfate calculated in the previous step to find the required volume of the solution. Volume of sodium thiosulfate solution = \(\frac{\textrm{Moles of sodium thiosulfate}}{\textrm{Molarity}} = \frac{0.197 \textrm{mol}}{0.244 \textrm{M}} = 0.807 \textrm{L}\) The required volume of 0.244 M sodium thiosulfate solution to react completely with 25.0 g of iodine is 0.807 L.

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Most popular questions from this chapter

I A student is given \(0.930 \mathrm{~g}\) of an unknown acid, which can be either oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), or citric acid, \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\). To determine which acid she has, she titrates the unknown acid with \(0.615 \mathrm{M} \mathrm{NaOH}\). The equivalence point is reached when \(33.6 \mathrm{~mL}\) are added. What is the unknown acid?

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) are combined, the following reaction occurs: $$ 2 \mathrm{PO}_{4}{ }^{3-}(a q)+3 \mathrm{Ca}^{2+}(a q) \longrightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) $$ How many grams of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)(\mathrm{MM}=310.18 \mathrm{~g} / \mathrm{mol})\) are obtained when \(15.00 \mathrm{~mL}\) of \(0.1386 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) are mixed with \(20.00 \mathrm{~mL}\) of \(0.2118 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Classify each of the following half-reactions as oxidation or reduction. (a) \(\mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{3+}(a q)\) (b) \(\mathrm{Zn}^{2+}(a q) \longrightarrow \mathrm{Zn}(s)\) (c) \(\mathrm{NH}_{4}{ }^{+}(a q) \longrightarrow \mathrm{N}_{2}(g)\) (d) \(\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{CH}_{2} \mathrm{O}(a q)\)

Using circles to represent cations and squares to represent anions, show pictorially the reactions that occur between aqueous solutions of (a) \(\mathrm{Ba}^{2+}\) and \(\mathrm{OH}^{-}\) (b) \(\mathrm{Co}^{3+}\) and \(\mathrm{PO}_{4}^{3-}\)

III Classify each of the following half-reactions as oxidation or reduction. (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\)
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