Question

Sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\right),\) the major component in photographic fixer solution, reacts with silver bromide to dissolve it according to the following reaction: \(\mathrm{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q) \longrightarrow\) $$ \mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}(a q)+\mathrm{NaBr}(a q) $$ (a) How many moles of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) would be required to react completely with \(0.450 \mathrm{~g}\) of \(\mathrm{AgBr}\) ? (b) How many \(\mathrm{mL}\) of \(0.02 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) contain this number of moles?

Short answer

0.004790 mol of Na2S2O3 are needed, requiring 239.5 mL of 0.02 M solution.

Step-by-step solution

Calculate Molar Mass of AgBr

To find the number of moles of AgBr, first calculate its molar mass. The molar mass of silver (Ag) is approximately 107.87 g/mol, and for bromine (Br) it is approximately 79.90 g/mol. Thus, the molar mass of AgBr is: \[ M(\mathrm{AgBr}) = 107.87 + 79.90 = 187.77 \text{ g/mol} \]

Calculate Moles of AgBr

Using the molar mass calculated in Step 1, determine the number of moles of AgBr in 0.450 g:\[ n(\mathrm{AgBr}) = \frac{0.450 \text{ g}}{187.77 \text{ g/mol}} = 0.002395 \text{ mol} \]

Determine Moles of Na2S2O3 Required

From the balanced chemical equation, one mole of AgBr reacts with two moles of \( \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \). Thus, the moles of \( \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \) needed:\[ n(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}) = 2 \times n(\mathrm{AgBr}) = 2 \times 0.002395 = 0.004790 \text{ mol} \]

Calculate Volume of Na2S2O3 Solution

To find how many milliliters of a 0.02 M \( \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \) solution are required, use the molarity formula: \[ M = \frac{n}{V} \] where \( V \) is in liters. Rearrange to find \( V \):\[ V = \frac{n}{M} = \frac{0.004790 \text{ mol}}{0.02 \text{ mol/L}} = 0.2395 \text{ L} \]Convert liters to milliliters:\[ 0.2395 \text{ L} = 239.5 \text{ mL} \]

Key concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that describes the mass of one mole of a substance. This calculation is crucial for converting between mass and moles, which helps us in solving various stoichiometric problems. To calculate the molar mass, we sum up the atomic masses of all the elements present in the formula. For example, in the case of silver bromide (AgBr):

• Atomic mass of Silver (Ag) is about 107.87 g/mol.
• Atomic mass of Bromine (Br) is about 79.90 g/mol.

Thus, the molar mass of AgBr is the total of these values, which is 187.77 g/mol. Understanding and accurately performing molar mass calculations allow chemists to convert grams to moles, which is the first step in any stoichiometric analysis.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through breaking and forming bonds. In the given exercise, we focus on the interaction between sodium thiosulfate (\(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\)) and silver bromide (\(\mathrm{AgBr}\)). This reaction dissolves AgBr in a photographic fixer solution:

• The balanced chemical equation shows that one mole of AgBr reacts with two moles of Na₂S₂O₃.
• The products formed are Na₃Ag(S₂O₃)₂ and NaBr.

Understanding such equations helps chemists predict the amount of reactants needed and the quantity of products formed in a reaction.

Solution Concentration

Solution concentration, often expressed as molarity, indicates the amount of a solute present in a unit volume of solution. Molarity (M) is defined as moles of solute per liter of solution. In a reaction, knowing the concentration allows for the calculation of how much of a solution is required to provide a specific number of moles of a reactant or product. For example, a 0.02 M (\(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\)) solution means there are 0.02 moles of Na₂S₂O₃ per liter of solution. By using the molarity formula (\( M = \frac{n}{V} \)), one can find the volume of solution needed by rearranging to solve for V, which gives control over how much reagent is used in a chemical reaction.

Volumetric Conversions

In chemistry, converting between different units of measurement is essential, especially when dealing with liquid volumes. After obtaining the necessary volume in liters from the molarity equation, it often needs to be converted to milliliters since laboratory measurements in milliliters are more practical. The conversion is straightforward:

• 1 liter (L) is equal to 1000 milliliters (mL).

For instance, if you've calculated that you need 0.2395 liters of a solution, multiply this value by 1000 to convert it to milliliters, resulting in 239.5 mL. Such conversions ensure precision in experiments and are vital for accurate quantitative analysis.