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Chapter 16: Q12P (page 391)

From information in Table 16-3, explain how you would use ${KMnO}_{4}$ to find the content of ${({NH}_{4})}_{2} S_{2} O_{8}$ in a solid mixture with ${({NH}_{4})}_{2} S_{2} O_{4}$ What is the purpose of phosphoric acid in the procedure?

Short Answer

Expert verified

The end point of phosphoric acid is the development of yellow colour in ${Fe}^{3 +}$

Step by step solution

01

Concept used

The answer is not given in the Drive.

02

How to determine the content of (NH4)S2O8x using potassium permanganate in the solid mixture with (NH4)S2O4 must be specified, as well as the function of phosphoric acid in this operation.

Given data: For $S_{2} O_{8}^{2 -}$ species, the reaction is

Given as,

$S_{2} O_{8}^{2 -} + 2 {Fe}^{2 +} + 2 H^{+} 鈬� 2 {Fe}^{3 +} 2 {HSO}_{4}^{-}$

Extra standard is supplemented with peroxydisulphate.

${Fe}^{2 +}$ having $H_{3} {PO}_{4}$ Remaining ${Fe}^{2 +}$ titrated with ${MnO}_{4}^{-}$

A weighted amount of the solid combination is added to the solution containing more standard ${Fe}^{2 +}$ and phosphoric acid.

Each mol $({NH}_{4}) S_{2} O_{8}$ of oxidises two mol of ${Fe}^{2 +}$ and ${Fe}^{3 +}$

To compute the quantity of ${Fe}^{2 +}$ consumed by $({NH}_{4}) S_{2} O_{8}$ extra ${Fe}^{2 +}$ is titrated with normal potassium permanganate.

The end point of phosphoric acid is the development of yellow colour in ${Fe}^{3 +}$

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Most popular questions from this chapter

Some people have an allergic reaction to the food preservative sulfite $(S O_{3}^{2 -})$ , 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 $(0.8043 {gKIO}_{3} + 6.0 gKI) / 100 mL$ . Acidification with 1.0 mL of $6.0 M H_{2} S O_{4}$ quantitatively converted role="math" localid="1663606948648" $l O_{3}$ into $l_{3}$ . The $l_{3}$ reacted with $S O_{3}^{2 -}$ to generate role="math" localid="1663607055826" $S O_{4}^{2 -}$ , leaving excess $l_{3}$ in solution. The excess $l_{3}$ required of $12.86 m L o f 0.04818 M N a_{2} S_{2} O_{3}$ to reach a starch end point.

(a) Write the reaction that occurs when $H_{2} S O_{4}$ is added to $K I O_{3} +$ 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 $H_{2} S O_{4}$ accurately?

(b) Write a balanced reaction between $l_{3}$ and sulfite.

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

(d) t test. Another wine was found to contain $277.7 m g S O_{3}^{2 -} / L$ with a standard deviation of $卤 2.2 m g / L$ for three determinations by the iodimetric method. A spectrophotometric method gaverole="math" localid="1663607422230" $273.2 卤 2.1 m g / L$ in three determinations. Are these results significantly different at the 95 % confidence level?

(a) Potassium iodate solution was prepared by dissolving 1.022gof $K I O_{3} (F M 214.00)$ in a 500 - mLvolumetric flask. Then 50.00mL of the solution were pipetted into a flask and treated with excess KI (2g) and acid $(10 m L o f 0.5 M H_{2} S O_{4})$ ofHow many moles of $f l_{3}^{-}$ are created by the reaction?

(b) The triiodide from part (a) reacted with 37.66 mL of $N a_{2} S_{2} O_{3}$ solution. What is the concentration of the $N a_{2} S_{2} O_{3}$ solution?

(c) A 1.223-g sample of solid containing ascorbic acid and inert ingredients was dissolved in dilute $H_{2} S O_{4}$ and treated with 2g of KI and 50.00mL of $K I O_{3}$ solution from part (a). Excess triiodide required14.22 mLof $N a_{2} S_{2} O_{3}$ solution from part (b). Find the weight percent of ascorbic acid (FM 176.13) in the unknown.

(d) Does it matter whether starch indicator is added at the beginning or near the end point in the titration in part (c)?

Two possible reactions of ${MnO}_{4}^{-}$ with $H_{2} O_{2}$ to produce $O_{2}$ andare ${Mn}^{+}$

Scheme: ${MnO}_{4}^{-} 鈫� {Mn}^{2 +} H_{2} O_{2} 鈫� O_{2}$

Scheme: ${MnO}_{4}^{-} 鈫� O_{2} + {Mn}^{2 +} H_{2} O_{2} 鈫� H_{2} O$

(a) Complete the half reactions for both schemes by adding $e^{+}$ and $H_{2} O$ and $H^{+}$ write a balanced net equation for each scheme.

(b) Sodium peroxyborate tetrahydrate, ${NaBO}_{3} 鈰� 4 H_{2} O (FM 153.86)$ produces $H_{2} O_{2}$ when dissolved in acid ${BO}_{3}^{-} + 2 H_{2} O 鈫� H_{2} O_{2} + H_{2} {BO}_{3}^{-}$ . To decide whether Scheme 1 or 2 Schemeoccurs student at the U.S. Naval academy weighed $0.123 {gNaBO}_{3} . 2 H_{2} O$ into a 100mLvolumetric flask added 20mLof $1 M H_{2} {SO}_{4}$ and diluted to the mark with $H_{2} O$ . Then they titratedof this solution with $0.01046 {MKMnO}_{4}$ until the first pale pink color persisted. How may mL of ${KMnO}_{4}$ are required in Scheme 1and 2 Scheme?

(The Scheme 1stoichiometry was observed).

From the following reduction potentials

$l_{2} (s) + 2 e^{-} 鈬� 2 l E^{掳} = 0.535 V l_{2} (a q) + 2 c^{-} 鈬� 2 l^{-} E^{掳} = 0.620 V l_{3} + 2 e^{-} 鈬� 3 l 3 l E^{掳} = 0.535 V$

(a) Calculate the equilibrium constant for $l_{2} (aq) + l^{-} 鈬� l_{3}^{-} .$

(b)The equilibrium constant for $l_{2} (s) + l^{-} 鈬� l_{3}^{-} .$

(c)The solubility (g/L.) of $l_{2} (s)$ in water is.

When 25.00mLof unknown were passed through a Jones reductor, molybdate ion $(M o O_{4}^{2 -})$ was converted into. The filtrate required 16.43mLof $0.01033 {MKMnO}_{4}$ to reach the purple end point.

role="math" localid="1663608295687" ${MnO}_{4}^{-} + {Mo}^{3 +} 鈫� {Mn}^{2 +} + {MoO}_{2}^{2 +}$

A blank required. Balance the reaction and find the molarity of Molybdate in the unknown.

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