Problem 63 - Chemistry and Chemical Reactivity

Question

In the photographic developing process, silver bromide is dissolved by adding sodium thiosulfate. \(\mathrm{AgBr}(\mathrm{s})+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow\) $$ \mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}(\mathrm{aq})+\mathrm{NaBr}(\mathrm{aq}) $$ If you want to dissolve \(0.225 \mathrm{g}\) of \(\mathrm{AgBr}\), what volume of \(0.0138 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3},\) in milliliters, should be used? (IMAGE CANNOT COPY)

Step-by-Step Solution

Answer

173.7 mL of 0.0138 M Na2S2O3 is needed.

Step 1: Calculate Moles of AgBr

First, find the number of moles of AgBr using the 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 of Reaction

From the balanced chemical equation, we know that 1 mole of \(\mathrm{AgBr}\) reacts with 2 moles of \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\). Therefore, double the moles of \(\mathrm{AgBr}\) to find the moles of \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\) needed:\[\text{moles of Na}_2\mathrm{S}_2\mathrm{O}_3 = 2 \times 0.001198\, \text{mol} = 0.002396\, \text{mol}\]

Step 3: Find Volume of Na2S2O3 Solution

Using the molarity equation, \(\mathrm{M} = \frac{\text{moles}}{\text{volume (in L)}}\), solve for volume:\[\text{Volume} = \frac{\text{moles}}{\text{Molarity}} = \frac{0.002396\, \text{mol}}{0.0138\, \mathrm{M}} \approx 0.1737\, \text{L}\]Convert liters to milliliters:\[0.1737\, \text{L} \times 1000 \frac{\text{mL}}{\text{L}} = 173.7\, \text{mL}\]

Key Concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that involves calculating the relationships between reactants and products in chemical reactions. It's like a recipe that tells you how much of each ingredient you need. Here’s how it’s typically broken down:

When you have a balanced chemical equation, it helps you understand the proportions of each substance involved. For example, in our exercise, the equation is:

• \( \mathrm{AgBr} (\mathrm{s}) + 2 \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3(\mathrm{aq}) \rightarrow \mathrm{Na}_3 \mathrm{Ag}\left(\mathrm{S}_2 \mathrm{O}_3 \right)_2(\mathrm{aq}) + \mathrm{NaBr}(\mathrm{aq}) \)

This equation indicates that 1 mole of silver bromide ( \( \mathrm{AgBr} \)) reacts with 2 moles of sodium thiosulfate ( \( \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \)). This balanced equation is crucial because if you know the amount of one substance, you can calculate the amount of another substance needed or produced.

Stoichiometry is based on the law of conservation of mass, meaning everything that goes into the reaction comes out, just in a different form. By using molar ratios derived from the balanced equation, we ensure that the conservation of mass is maintained during our calculations.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. They are represented by chemical equations, like the one seen in our exercise. These equations not only show you what's changing, but also how it's changing.

When looking at the chemical reaction in the photographic developing process:

• The reactants are silver bromide and sodium thiosulfate.
• The products are sodium silver thiosulfate and sodium bromide.

Reactions tell us the chemical substances we start with and the new substances we end up with after the change. Each component of the reaction has coefficients in front of them, indicating how many moles of each substance participate in the reaction.

In our case, the coefficients indicate that one mole of silver bromide combines with two moles of sodium thiosulfate. Understanding these coefficients allows you to determine the stoichiometric relationships needed for calculations, such as determining how much reagent is required to complete a reaction.

Molarity Calculations

Molarity is an essential concept when dealing with solutions in chemistry. It is defined as the number of moles of solute per liter of solution (mol/L) and is given by the formula:

• \( \text{M} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

For the exercise, we calculated how much sodium thiosulfate was required to dissolve a given amount of silver bromide. After determining the moles of \( \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \) needed to react with the silver bromide, we used molarity to find out the volume of solution required.

The calculated moles of sodium thiosulfate were divided by the given molarity (0.0138 M) to find the volume in liters. This volume was then converted to milliliters for practical use, as solutions in labs are often measured in milliliters.

This application of molarity allows chemists to precisely control the concentrations and volumes of solutions they use, ensuring reactions proceed as desired and products are correctly formed.