Question

Some people have an allergic reaction to the food preservative sulfite $\left(S{O}_3^{2-}\right)$, which can be measured by instrumental methods 37 or by a redox titration: To 50.0mL of wine were added 50.0mL of solution containing $\left(0.8043{\mathrm{gKIO}}_3+6.0\mathrm{gKI}\right)\mathbf{/}\mathbf{100}\mathbf{mL}$ . Acidification with 1.0 mL of $6.0M{H}_{2}S{O}_4$ quantitatively converted role="math" localid="1663606948648" $\mathit{l}{\mathit{O}}_{\mathbf{3}}$ into ${\mathit{l}}_{\mathbf{3}}$ . The ${\mathit{l}}_{\mathbf{3}}$ reacted with $\mathit{S}{\mathit{O}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$ to generate role="math" localid="1663607055826" $\mathit{S}{\mathit{O}}_4^{\mathbf{2}\mathbf{-}}$ , leaving excess ${\mathit{l}}_{\mathbf{3}}$ in solution. The excess ${\mathit{l}}_{\mathbf{3}}$ required of $\mathbf{12}\mathbf{.}\mathbf{86}\mathbf{}\mathit{m}\mathit{L}\mathbf{}\mathit{o}\mathit{f}\mathbf{0}\mathbf{.}\mathbf{04818}\mathbf{}\mathit{M}\mathit{N}{\mathit{a}}_{\mathbf{2}}{\mathit{S}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$to reach a starch end point.

(a) Write the reaction that occurs when ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$is added to $\mathit{K}\mathit{I}{\mathit{O}}_{\mathbf{3}}\mathbf{+}$ KI and explain why 6.0 gKI were added to the stock solution. Is it necessary to measure out 6.0 g accurately? Is it necessary to measure 1.0 mL
of${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$ accurately?

(b) Write a balanced reaction between${\mathit{l}}_{\mathbf{3}}$ and sulfite.

(c) Find the concentration of sulfite in the wine. Express your answer in mol/L and in$\mathit{m}\mathit{g}\mathit{S}{\mathit{O}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$ per liter.

(d) t test. Another wine was found to contain $\mathbf{277}\mathbf{.}\mathbf{7}\mathit{m}\mathit{g}\mathbf{}\mathit{S}{\mathit{O}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{/}\mathit{L}$with a standard deviation of $\mathbf{\pm }\mathbf{2}\mathbf{.}\mathbf{2}\mathit{m}\mathit{g}\mathbf{/}\mathit{L}$for three determinations by the iodimetric method. A spectrophotometric method gaverole="math" localid="1663607422230" $\mathbf{273}\mathbf{.}\mathbf{2}\mathbf{\pm }\mathbf{2}\mathbf{.}\mathbf{1}\mathit{m}\mathit{g}\mathbf{/}\mathit{L}$ in three determinations. Are these results significantly different at the 95 % confidence level?

Short answer

(a) The reaction requires a large amount of sulphuric acid, which is in excess. Sulphuric acid and potassium iodide do not need to be precisely measured.

(b) The balanced reaction of triiodide and sulphate is,$I_3+S{O}_3^{2-}+H_{2}O\rightleftharpoons 31+S{O}_4^{2-}+2H^{+}$

(c) The amount of sulphite discovered is $\left(80.06\right)\times \left(5.079\right)=406.6mg/L$

(d) The difference is not significant at the 95 percent confidence level.

Step-by-step solution

Definition for conversion of mass in a chemical reaction and Molarity

• There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
• First write the reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Finally obtained values could place it as coefficients of reactants as well as products.
• Molarity (M): One of the concentration units is molarity. The number of moles of the solute in one litre of solution is known as molarity.
• Molarity $=\frac{Numberofmolesofthesolute}{IL.solution}$

Determine, is it necessary to measure 1.0 mL  of  H2SO4 accurately

(a)

When sulphuric acid is added to water, a reaction happens.$KI{O}_3+KI$ It must be written, and the purpose for adding potassium iodide to the stock solution must be clarified. It is critical to take accurate measurements$6.0g$and$1.0mL$of$H_{2}S{O}_4$ It is necessary to provide precise information.

To write: the reaction that occurs when Sulphuric acid is mixed with $KI{O}_3+KI$

$I{O}_3^{-}+8I^{-}+6H^{+}\rightleftharpoons 3I_3+3H_{2}O$

To elucidate: why was potassium iodide added to the stock solution?

Having a stock solution$\left(0.8403gKI{O}_3+5gKI\right)/100mL$.

This solution translates to $0.03758MKI{O}_3$and $0.18792mmolKI{O}_3+1.5mmolKI$.

To give: Measuring out is critical 6.0 g and 1.0 ml of $H_{2}S{O}_4$exactly

1.0 mL of $6.0M{H}_{2}S{O}_4$has 6mmol The reaction requires a large amount of sulphuric acid, which is in excess. Sulphuric acid and potassium iodide do not need to be precisely measured.

Determine the balanced reaction between I3  and sulfite

(b)

The balanced reaction between triiodide and sulphate has to be written.

To write: a well-balanced triiodide-sulphate reaction

Lodine and sulphide are formed when triiodide and sulphate combine. Triiodide is reduced, and sulphate is oxidised in this redox process. As a result, the balanced response is,

$I_3+S{O}_3^{2-}+H_{2}O\rightleftharpoons 3I+S{O}_4^{2-}+2H^{+}$

Determine the concentration of sulfite in the wine

(c)

The sulphite concentration in the wine must be calculated. The solution should be given in both mol/L and mol/L$mgS{O}_3^{2-}/L$

The sulphite concentration in the wine is found to be$5.079\times {10}^{-3}M$ The amount of sulphite in the mixture is$406.6mg/L$.

To calculate: the sulphite content of the wine $0.18792mmolKI{O}_3$when brought to the wine $3\times 0.18792=0.56376mmol{l}_3^{-}$. An excess of unreacted triiodide is required $12.86mLof0.04818MN{a}_2{S}_2{O}_3=0.61959mmol$.

There should be two moles of sodium thiosulphite for every mole of unreacted triiodide (0.61959)/2 = 0.3098mmol The reaction with sulphite produced triiodide. As a result, triiodide reacts with sulphate $\left(0.56376-0.3098\right)=0.2540mmol{l}_3.1$ mol triiodide interacts with mol triiodide As a result, sulphite must have been present $0.2540mmolS{O}_3^{2-}in50.0mL$ of wine.

Concentration of sulphite $=\frac{0.2540mmol}{50.0mL}=5.079\times {10}^{-3}M$

Sulphite has a mass formula of 80.06. The amount of sulphite discovered is $\left(80.06\right)\times \left(5.079\right)=406.6mg/L$

Determine the results significantly different at the 95 % confidence level

(d)

It must be determined whether the outcomes stated in the statement are significantly different at the 95% confidence level.$$\begin{gathered} s_{pooled}=\sqrt{\frac{2.2^2\left(3-1\right)+2.1^2\left(3-1\right)}{3+3-2}}=2.15 \\ {t}_{calculated}=\frac{277.7-273.21}{215}\sqrt{\frac{3.3}{3+3}}=2.56 \end{gathered}$$

To clarify: Is there a 95 percent chance that the outcomes in the statement are significantly different $t_{calulated}=2.776$for 95 % 3+3-2=4 and confidence As a result, there are four degrees of freedom $t_{calulated}<t_{table}$, As a result, the difference is not significant at the 95 percent confidence level.