Question

In the photographic developing process, silver bromine is dissolved by adding sodium thiosulphate: $$ \begin{array}{r} \mathrm{AgBr}(\mathrm{s})+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \\ \quad \mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}(\mathrm{aq})+\mathrm{NaBr}(\mathrm{aq}) \end{array} $$ If you want to dissolve \(0.250 \mathrm{~g}\) of \(\mathrm{AgBr}\), what volume of \(0.0138 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\), in milliliters, should be used? a. \(143 \mathrm{ml}\) b. \(139 \mathrm{ml}\) c. \(96.5 \mathrm{ml}\) d. \(193 \mathrm{ml}\)

Short answer

The required volume is 193 mL (option d).

Step-by-step solution

Find Molar Mass of AgBr

Calculate the molar mass of \(\mathrm{AgBr}\). Silver (Ag) has a molar mass of \(107.87\, \text{g/mol}\), and bromine (Br) has a molar mass of \(79.90\, \text{g/mol}\). Adding these values together, the molar mass of \(\mathrm{AgBr}\) is:\[ \text{Molar mass of } \mathrm{AgBr} = 107.87 + 79.90 = 187.77\, \text{g/mol} \]

Calculate Moles of AgBr

Using the molar mass of \(\mathrm{AgBr}\) from Step 1, find the moles of \(\mathrm{AgBr}\) in \(0.250\, \text{g}\).\[ \text{Moles of } \mathrm{AgBr} = \frac{0.250\, \text{g}}{187.77\, \text{g/mol}} \approx 1.33 \times 10^{-3}\, \text{moles} \]

Determine Moles of Na2S2O3 Needed

Using the balanced chemical equation:\[ \mathrm{AgBr} + 2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \rightarrow \mathrm{Na}_{3} \mathrm{Ag}(\mathrm{S}_{2} \mathrm{O}_{3})_{2} + \mathrm{NaBr} \] Note that 1 mole of \(\mathrm{AgBr}\) reacts with 2 moles of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\). Therefore, multiply the moles of \(\mathrm{AgBr}\) by 2 to find the moles of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\):\[ \text{Moles of } \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} = 2 \times 1.33 \times 10^{-3} = 2.66 \times 10^{-3}\, \text{moles} \]

Convert Moles of Na2S2O3 to Volume in Liters

Using the molarity (concentration) of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\), calculate the volume in liters needed to provide the \(2.66 \times 10^{-3}\, \text{moles}\):\[ \text{Volume in liters} = \frac{2.66 \times 10^{-3}\, \text{moles}}{0.0138\, \text{M}} = 0.193\, \text{L} \]

Convert Volume to Milliliters

Convert the volume from liters to milliliters:\[ \text{Volume in milliliters} = 0.193\, \text{L} \times 1000\, \frac{\text{mL}}{\text{L}} = 193\, \text{mL} \]

Key concepts

Silver Bromide Dissolution

Silver bromide, denoted as \( \text{AgBr} \), is a chemical compound used in photographic film. It needs to be dissolved during the developing process to reveal the captured image. In this process, sodium thiosulphate plays a crucial role as a fixing agent. It reacts with silver bromide to dissolve it and form soluble complexes, allowing the film to be developed without further exposure to light.

The chemical reaction involved is:\[ \text{AgBr} + 2 \text{Na}_2 \text{S}_2 \text{O}_3 \rightarrow \text{Na}_3 \text{Ag(S}_2 \text{O}_3)_2 + \text{NaBr} \]
\( \text{AgBr} \) reacts with sodium thiosulphate \( \text{Na}_2 \text{S}_2 \text{O}_3 \) to form a soluble silver thiosulphate complex and sodium bromide \( \text{NaBr} \), both of which are soluble in water, thus removing the unreacted silver bromide from the film.

Without this dissolution process, the unreacted \( \text{AgBr} \) would remain on the film, leading to undeveloped areas and a ruined photograph. This step ensures that only the areas exposed to light, which form actual silver, are retained on the film.

Chemical Reaction Stoichiometry

Understanding chemical reaction stoichiometry is key to determining how much of each reactant is needed for a reaction to proceed completely. The stoichiometry of a reaction refers to the quantitative relationships between the amounts of reactants and products in a chemical reaction.

In the reaction of silver bromide with sodium thiosulphate:

• The balanced equation is: \( \text{AgBr} + 2 \text{Na}_2 \text{S}_2 \text{O}_3 \rightarrow \text{Na}_3 \text{Ag(S}_2 \text{O}_3)_2 + \text{NaBr} \).
• This tells us that for every mole of \( \text{AgBr} \), two moles of \( \text{Na}_2 \text{S}_2 \text{O}_3 \) are required.

The stoichiometric coefficients (numbers in front of the compounds) show the ratio of reactants needed and products formed.

This ratio guides you in calculating the exact amount of each substance you will need in a reaction. In this exercise, we used stoichiometry to determine the moles of \( \text{Na}_2 \text{S}_2 \text{O}_3 \) needed, ensuring the \( \text{AgBr} \) dissolves completely.

Molar Mass Calculation

Molar mass is a vital concept that connects the mass of a substance to the amount of substance present in moles, providing the conversion factor necessary to relate mass to moles. Calculating the molar mass involves summing up the atomic masses of each element present in a compound, following the chemical formula.

For silver bromide \( \text{AgBr} \):

• The atomic mass of silver (\( \text{Ag} \)) is approximately \( 107.87 \, \text{g/mol} \).
• The atomic mass of bromine (\( \text{Br} \)) is approximately \( 79.90 \, \text{g/mol} \).

Adding these gives the molar mass:\[ \text{Molar mass of } \mathrm{AgBr} = 107.87 + 79.90 = 187.77 \, \text{g/mol} \]

Once you have the molar mass, you can convert between mass and moles by using the formula:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} \]
This allows you to determine how many moles of a substance you have, which is crucial when using stoichiometry to find out how much of another substance you will need to react completely. In our case, it helped to find out the moles of \( \text{AgBr} \) in \( 0.250 \, \text{g} \) for the photographic process.