Problem 111 - Chemistry: An Atoms-Focused Approach

Question

To determine the concentration of \(\mathrm{SO}_{4}^{2-}\) ion in a sample of groundwater, \(100.0 \mathrm{mL}\) of the sample is titrated with \(0.0250 M \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2},\) forming insoluble \(\mathrm{BaSO}_{4} .\) If \(3.19 \mathrm{mL}\) of the \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution is required to reach the end point of the titration, what is the molarity of the \(\mathrm{SO}_{4}^{2-} ?\)

Step-by-Step Solution

Answer

Answer: The molarity of the sulfate ions (SO4^2-) in the groundwater sample is 7.975 × 10^{-4} M.

Step 1: Write the balanced chemical equation for the reaction

The balanced chemical equation for the precipitation reaction between barium ions (Ba^2+) and sulfate ions (SO4^2-) is given by: Ba^2+ (aq) + SO4^2- (aq) -> BaSO4 (s)

Step 2: Calculate the moles of barium ions (Ba^2+) needed to reach the end point of the titration

The volume of barium nitrate solution required to reach the end point of the titration is 3.19 mL or 0.00319 L. Given the molarity of the barium nitrate solution (0.0250 M), we can calculate the moles of barium ions (Ba^2+) needed as follows: moles of Ba^2+ = volume × molarity moles of Ba^2+ = 0.00319 L × 0.0250 mol/L moles of Ba^2+ = 7.975 × 10^{-5} mol

Step 3: Use stoichiometry to find the moles of sulfate ions (SO4^2-) in the sample

From the balanced chemical equation, the mole ratio between barium ions (Ba^2+) and sulfate ions (SO4^2-) is 1:1. Therefore, the moles of sulfate ions in the sample will be equal to the moles of barium ions: moles of SO4^2- = moles of Ba^2+ moles of SO4^2- = 7.975 × 10^{-5} mol

Step 4: Calculate the molarity of the sulfate ions in the sample

The molarity of the sulfate ions can be calculated by dividing the moles of SO4^2- by the volume of the groundwater sample in liters: molarity of SO4^2- = moles of SO4^2- ÷ volume of sample molarity of SO4^2- = 7.975 × 10^{-5} mol ÷ 0.100 L molarity of SO4^2- = 7.975 × 10^{-4} M Therefore, the molarity of the sulfate ions (SO4^2-) in the groundwater sample is 7.975 × 10^{-4} M.

Key Concepts

Molarity

Molarity is a key concept in solution chemistry. It refers to the number of moles of a solute present in one liter of solution. It's expressed in moles per liter (mol/L) and is crucial for calculating concentrations and performing stoichiometric calculations. This measurement allows chemists to determine how much of a substance is available to react.

When carrying out a titration, knowing the molarity of a solution helps in figuring out the exact amount needed to react with a given quantity of another substance. For example, if you have a solution with a molarity of 0.0250 M, it means there are 0.0250 moles of the solute in each liter of the solution. This information is used in the calculations to find out how much solute reacts with a known quantity of another reactant. By knowing the volume of the solution used and its molarity, you can figure out the number of moles involved in the reaction.

Chemical Equation

A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants on the left and products on the right. A balanced chemical equation follows the law of conservation of mass, meaning the number of atoms for each element must be the same on both sides of the equation.

In the titration exercise, the balanced chemical equation shows the reaction of barium ions \((Ba^{2+})\) and sulfate ions \((SO_4^{2-})\) forming barium sulfate \((BaSO_4)\). The balanced reaction is \(\mathrm{Ba}^{2+} \mathrm{(aq)} + \mathrm{SO}_{4}^{2-} \mathrm{(aq)} \rightarrow \mathrm{BaSO}_4 \mathrm{(s)}\).
This equation helps determine the mole ratios of reactants and products. In this case, the mole ratio is 1:1, meaning one mole of barium ions reacts with one mole of sulfate ions to form barium sulfate.

Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It involves using the balanced chemical equation to determine the amounts of substances consumed and produced. This helps predict how much product will form from a given amount of reactant or vice versa.

In this example, the stoichiometry of the reaction tells us that the moles of barium ions \((Ba^{2+})\) used in the reaction equal the moles of sulfate ions \((SO_4^{2-})\) present in the solution because of the 1:1 molar ratio. Using the known amount of barium nitrate solution required to complete the reaction and its molarity, we can calculate the moles of sulfate ions. This helps find the concentration of sulfate ions in the original groundwater sample.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts react in solution to form an insoluble solid called a precipitate. This type of reaction can be used to determine concentrations of ions in a solution by observing the formation of a precipitate.

In this titration exercise, a precipitation reaction takes place. When barium nitrate solution \((Ba(NO_3)_2)\) is added to a solution containing sulfate ions \((SO_4^{2-})\), an insoluble solid, barium sulfate \((BaSO_4)\), forms. The appearance of this solid marks the endpoint of the titration. We use the volume of barium nitrate added to reach this endpoint to calculate the original concentration of sulfate ions in the groundwater.

• This method is particularly useful for accurately measuring concentrations of one reactant when the product forms a visible solid.
• The precipitation reaction in this context helps understand the purity and composition of natural water samples.