Problem 72 - Chemistry Matter and Change

Question

Film Photographic film contains silver bromide in gelatin. Once exposed, some of the silver bromide decomposes, producing fine grains of silver. The unexposed silver bromide is removed by treating the film with sodium thiosulfate. Soluble sodium silver thiosulfate \(\left(\mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right)\) is produced. \begin{equation} \mathrm{AgBr}(\mathrm{s})+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \end{equation} \begin{equation} \quad\quad\quad\quad\quad\quad\quad\mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}(\mathrm{aq})+\mathrm{NaBr}(\mathrm{aq}) \end{equation} Determine the mass of \(\mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\) produced if 0.275 \(\mathrm{g}\) of \(\mathrm{AgBr}\) is removed.

Step-by-Step Solution

Answer

Approximately 0.588 g of Na3Ag(S2O3)2 is produced.

Step 1: Find Molar Mass of AgBr

First, we need to find the molar mass of AgBr to convert the given mass into moles. The atomic masses are approximately Ag = 107.87 g/mol and Br = 79.90 g/mol. Thus, the molar mass of AgBr is:\[ M(\text{AgBr}) = 107.87 + 79.90 = 187.77 \text{ g/mol} \]

Step 2: Convert Mass of AgBr to Moles

Using the molar mass of AgBr, convert the given mass (0.275 g) to moles:\[ \text{moles of AgBr} = \frac{0.275 \text{ g}}{187.77 \text{ g/mol}} \approx 0.001464 \text{ mol} \]

Step 3: Stoichiometry of the Reaction

According to the balanced chemical equation, 1 mole of AgBr produces 1 mole of Na_{3}Ag(S_{2}O_{3})_{2}. Therefore, the moles of Na_{3}Ag(S_{2}O_{3})_{2} produced are:\[ \text{moles of Na}_{3}\text{Ag}(S_{2}O_{3})_{2} = 0.001464 \text{ mol} \]

Step 4: Find Molar Mass of Na3Ag(S2O3)2

The molar mass of Na_{3}Ag(S_{2}O_{3})_{2} is calculated using the atomic masses: Na = 22.99 g/mol, Ag = 107.87 g/mol, S = 32.06 g/mol, O = 16.00 g/mol. Thus,\[ M(\text{Na}_{3}\text{Ag}(S_{2}O_{3})_{2}) = 3(22.99) + 107.87 + 2(32.06 + 2(16.00)) \approx 401.94 \text{ g/mol} \]

Step 5: Calculate Mass of Na3Ag(S2O3)2 Produced

Finally, convert the moles of Na_{3}Ag(S_{2}O_{3})_{2} back into mass:\[ \text{mass of Na}_{3}\text{Ag}(S_{2}O_{3})_{2} = 0.001464 \text{ mol} \times 401.94 \text{ g/mol} \approx 0.588 \text{ g} \]

Key Concepts

Chemical Reactions

In a chemical reaction, substances known as reactants interact to form new substances called products. During this process, chemical bonds are broken and new bonds are formed, causing a transformation of matter. Each chemical reaction has its own set of reactants and products, and is often represented by a chemical equation. In our exercise, the chemical equation for the decomposition of silver bromide (0AgBr0) showcases the transformation into a different compound through interactions with sodium thiosulfate (0Na_{2}S_{2}O_{3}0). This balanced equation reveals that the combination of reactants yields sodium silver thiosulfate (0Na_{3}Ag(S_{2}O_{3})_{2}0) and sodium bromide (0NaBr0) as products.

Molar Mass

Molar mass is a key concept in stoichiometry, the branch of chemistry that involves the calculation of reactants and products in chemical reactions. It represents the mass of one mole of a substance, expressed in grams per mole (g/mol). Understanding molar mass helps us convert between the mass of a substance and the amount in moles, which is crucial for quantitative chemical analysis. In the exercise, we first determined the molar mass of silver bromide (AgBr) by summing the atomic masses of silver (Ag) and bromine (Br). This allows us to convert the 0.275 grams of AgBr given in the exercise into moles, setting the stage for further stoichiometric calculations.

Photographic Chemistry

Photographic chemistry is centered on the chemical processes that occur during the development of photographs. Traditional photographic film contains silver halides, such as silver bromide, embedded in a gelatin base. When exposed to light, some silver bromide decomposes, causing a reaction that produces metallic silver grains which form the latent image. To develop the film, unexposed silver bromide must be removed. This is achieved by treating the film with a fixer, such as sodium thiosulfate, which dissolves the unaltered silver bromide and preserves the image. Our exercise highlights this process by focusing on the chemical transformation that occurs when sodium thiosulfate interacts with silver bromide.

Silver Bromide

Silver bromide (0AgBr0) is a light-sensitive compound that has been extensively used in photographic films. Its sensitivity to light is the core reason for its usage in photography. When exposed to light, it undergoes a decomposition reaction that yields metallic silver, creating the visible image on film. However, the unexposed silver bromide needs to be removed in the developing process to finalize the photograph. As described in the exercise, treating the film with sodium thiosulfate is the means by which unreacted silver bromide is dissolved and washed away, allowing the preserved image to be processed further. Understanding the role of silver bromide in this context is essential for grasping the underlying principles of traditional photographic chemistry.