Problem 65 - Chemistry and Chemical Reactivity

Question

In the photographic developing process, silver bromide is dissolved by adding sodium thiosulfate. $$\operatorname{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}(\mathrm{aq})+\mathrm{NaBr}(\mathrm{aq})$$ If you want to dissolve 0.225 g of AgBr, what volume of \(0.0138 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3},\) in milliliters, should be used?

Step-by-Step Solution

Answer

173.9 mL of 0.0138 M Na2S2O3 solution is needed.

Step 1: Find Moles of AgBr

Calculate the moles of AgBr using its molar mass. The molar mass of AgBr is approximately 187.77 g/mol. Use the formula:\[\text{moles of AgBr} = \frac{\text{mass of AgBr}}{\text{molar mass of AgBr}} = \frac{0.225 \text{ g}}{187.77 \text{ g/mol}} \approx 0.001198 \text{ mol}\]

Step 2: Use Stoichiometry to Find Moles of Na2S2O3

According to the balanced equation, 1 mole of AgBr reacts with 2 moles of \( \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} . \)Therefore, the moles of \( \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \) required:\[\text{moles of } \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} = 2 \times 0.001198 \approx 0.002396 \text{ mol} \]

Step 3: Calculate Volume of Na2S2O3 Solution

Determine the volume of \( \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \) solution needed using the molarity equation:\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]Rearranging for volume gives:\[\text{Volume in liters} = \frac{\text{moles of } \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}}{\text{Molarity}}\]Plug in the values:\[\text{Volume in liters} = \frac{0.002396}{0.0138} \approx 0.1739 \text{ L}\]

Step 4: Convert Volume to Milliliters

Convert the volume from liters to milliliters (1 L = 1000 mL):\[\text{Volume in milliliters} = 0.1739 \times 1000 \approx 173.9 \text{ mL}\]

Key Concepts

Molarity

Molarity is a key concept in chemistry that describes how concentrated a solution is. It is defined as the number of moles of solute (in this case, sodium thiosulfate, \( \text{Na}_2\text{S}_2\text{O}_3 \)) divided by the volume of the solution in liters. The unit of molarity is moles per liter, often expressed as M. Understanding molarity allows us to relate the amount of solute to the volume of solution it is found in, which is crucial in solving many chemistry problems.
Here's how to calculate molarity:

• First, determine the number of moles of the solute.
• Next, measure the volume of the solution in which the solute is dissolved, converting mills to liters if necessary.
• Finally, divide the moles of the solute by the volume of the solution in liters.

For example, if we need to use a 0.0138 M solution of \( \text{Na}_2\text{S}_2\text{O}_3 \), molarity helps us know exactly what volume to use to ensure the correct reaction occurs.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of bonds, producing new materials. In the given exercise, a chemical reaction occurs when silver bromide (\( \text{AgBr} \)) reacts with sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \)) to produce a new sodium complex and sodium bromide (\( \text{NaBr} \)).
During a chemical reaction, it's essential to balance the chemical equation to abide by the law of conservation of mass, which states that atoms are not created or destroyed. In this process:

• 1 mole of \( \text{AgBr} \) reacts with 2 moles of \( \text{Na}_2\text{S}_2\text{O}_3 \).
• This stoichiometric ratio provides the basis for all subsequent calculations about reactants and products.

Understanding these fundamentals ensures we can predict the amounts of products formed and reactants needed with accuracy.

Photographic Developing Process

The photographic developing process is an intriguing application of chemistry where light exposure and chemical reactions work hand in hand to create images. In traditional photography, the film contains silver halide crystals, like silver bromide (\( \text{AgBr} \)), which are sensitive to light.
When exposed to light, these crystals form a latent image. During developing, these crystals react with a developer, such as sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \)), often called 'hypo'. This reaction dissolves the unexposed silver halide crystals, leaving behind the actual image, which is comprised of reduced silver.

• This reaction is critical in removing unused silver halide particles to reveal the photograph.
• Proper mastering of this process allows for high-quality image development by accurately manipulating chemicals and reaction conditions.

Understanding the chemistry behind the process allows photographers to achieve optimal results and manipulate contrast, exposure, and other photographic qualities effectively.