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Chapter 27: Q6P (page 767)

Why is it less desirable to wash AgCl precipitate with aqueous NaNO3 than with HNO3 solution?

Short Answer

Expert verified

The AgCI precipitate ${HNO}_{3}$ would be more desirable than with ${NaNO}_{3}$

Step by step solution

01

Definition

It is an analytical technique that uses a precipitation reaction to separate ions from a solution. The chemical that is added to cause the precipitation is called the precipitant or precipitating agent

02

Characteristics of HNO3 and NANO3

doesn’t evaporate during drying.

evaporates during drying.

03

Washing of AgCl precipitate solution.

If we washed with AgCI precipitate with ${HNO}_{3},$ would evaporate during drying this would leave a clean AgCI precipitate.

If we washed with AgCI precipitate with ${NaNO}_{3}, {NaNO}_{3},$ is non volatile and wouldn’t evaporate during drying this would leave a dirty AgCI precipitate.

Therefore, precipitate with ${HNO}_{3}$ would more desirable than that ${NaNO}_{3}$ .

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Most popular questions from this chapter

1.475-g sample containing ${NH}_{4} Cl (FM 53.491), K_{2} {CO}_{3}$ (FM 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0-mL aliquot was acidified and treated with excess sodium tetraphenylborate, ${Na}^{+} B (C_{6} H_{5})_{4}^{-}$ , to precipitate $K^{+}$ and

${NH}_{4}^{+}$ ions completely:

$(C_{6} H_{5})_{4} B^{-} + K^{+} \to (C_{6} H_{5})_{4} B K (s) F M 358.33 (C_{6} H_{5})_{4} B^{-} + N H_{4}^{+} \to (C_{6} H_{5})_{4} B N H_{4} (s) F M 337.27$

The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original solution was made alkaline and heated to drive off all the ${NH}_{3} .$

$N H_{4}^{+} + O H^{-} \to N H_{3} (g) + H_{2} O$

It was then acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. Find the weight percent of ${NH}_{4} Cl$ and $K_{2} {CO}_{3}$ in the original solid.

Explain what is done in thermogravimetric analysis.

How many grams of AgBr would have been formed from 0.100 g of ${RaBr}_{2}$ ?

What is the purpose of the ${WO}_{3}$ and Cu in Figure 27-8?

Thermogravimetric analysis and propagation of error. 16 Crystals of deuteratedpotassium dihydrogen phosphate, $K {(D_{x} H_{1 - x})}_{2} {PO}_{4}$ are used in optics as a light valve, as a light deflector, and for frequency doubling of lasers. The optical properties are sensitive to the fraction of deuterium in the material. A publication states that deuterium content can be determined by measuring the mass lost by dehydration of the crystal after slow heating to $450 ° C$ in a Pt crucible under $N_{2}$

(a) Let $α$ be the mass of product divided by the mass of reactant:

$α = \frac{mass of {KPO}_{3}}{mass of K {(D_{x} H_{1 - x})}_{2} {PO}_{4}}$

Show that the coefficient xin $K {(D_{x} H_{1 - x})}_{2} {PO}_{4}$ is related to $α$ by the equation

$x = \frac{58.6670}{α} - 67.6183$

What would be the value of $α$ if starting material were 100% deuterated?

(b) One crystal measured in triplicate gave an average value of $α = 0 . 8567_{7}$ .Find x for this crystal.

(c) From equation B.1 in Appendix B, show that the uncertainty in $x (e_{x})$ is related to the uncertainty in $α (e_{α})$ by the equation

$e_{x} = \frac{58.6670 e_{α}}{α^{2}}$

(d) The uncertainty in deuterium: hydrogen stoichiometry is $e_{x}$ The authors estimate that their uncertainty in $α$ is $e_{x} = 0.0001 .$ From $e_{α}$ compute $e_{x}$ . Write the stoichiometry in the form $x \pm e_{x}$ If $e_{x}$ were 0.001 (which is perfectly reasonable), what would $e_{x}$ be?

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