Problem 97 - Chemistry

Question

Nitrate salts are generally considered to be soluble salts. One of the least soluble nitrate salts is barium nitrate. Approximately 15 g of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) will dissolve per liter of solution. Calculate the \(K_{\mathrm{sp}}\) value for barium nitrate.

Step-by-Step Solution

Answer

The \(K_{sp}\) value for barium nitrate is \(4.3 \times 10^{-5}\).

Step 1: Determine the molar solubility from the given solubility

Firstly, convert the solubility of barium nitrate in g/L to moles/L (molarity): Given solubility = 15 g/L Molar mass of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_2\) = 137.33 g/mol (Ba) + 2 * (14.01 g/mol (N) + 3 * 16.00 g/mol (O)) = 261.34 g/mol Therefore, molar solubility (S) = (15 g/L) / (261.34 g/mol) = 0.0574 mol/L

Step 2: Write the balanced reaction and dissociation expressions for barium nitrate

The dissociation of barium nitrate in water is represented by the following chemical equation: \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \rightleftharpoons \mathrm{Ba}^{2+} + 2\mathrm{NO}_{3}^{-}\) At equilibrium, the concentrations of \(\mathrm{Ba}^{2+}\) and \(\mathrm{NO}_{3}^-\) are [Ba2+] = S (since 1 mol of \(\mathrm{Ba}^{2+}\) is produced for every 1 mol of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) dissociated) and [NO3-] = 2S (since 2 mol of \(\mathrm{NO}_{3}^-\) are produced for every 1 mol of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) dissociated), The solubility product constant expression will be: \(K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{NO}_{3}^-]^2\)

Step 3: Calculate the \(K_{sp}\) value using the molar solubility

Now, we substitute the values of [Ba2+] and [NO3-] into the \(K_{sp}\) expression: \(K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{NO}_{3}^-]^2 = (S)(2S)^2\) Plugging in the molar solubility value (S = 0.0574 mol/L) to the equation: \(K_{sp} = (0.0574)(2\cdot0.0574)^2 = 4.3 \times 10^{-5}\) So the \(K_{sp}\) value for barium nitrate is \(4.3 \times 10^{-5}\).

Key Concepts

Molar Solubility

Molar solubility is a key concept when determining how much of a particular substance can dissolve in a solvent at a given temperature until the solution reaches saturation.
This is often expressed in units of moles per liter (mol/L).
In the case of barium nitrate, the process begins by converting its solubility from grams per liter (g/L) to molarity (mol/L).

• First, the given solubility is 15 g/L.
• The molar mass of barium nitrate (\(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\)) is calculated to be 261.34 g/mol.
• To find the molar solubility (\(S\)), the equation is: \[S = \frac{15 \text{ g/L}}{261.34 \text{ g/mol}} = 0.0574 \text{ mol/L}\]

This conversion allows us to use precise molarity values in further equilibrium calculations, such as those involving solubility product constants.

Barium Nitrate

Barium nitrate is a chemical compound with the formula \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\).
It consists of barium ions (\(\mathrm{Ba}^{2+}\)) and nitrate ions (\(\mathrm{NO}_3^−\)).

• Many nitrate salts are highly soluble in water, but barium nitrate is relatively less soluble compared to others.
• When it dissolves, it dissociates into one barium ion and two nitrate ions per formula unit.

This characteristic dissociation plays a significant role in determining the solubility product constant (\(K_{sp}\)), which quantifies a compound's solubility in water.

Solubility

Solubility is a measure of how much solute can dissolve in a solvent to form a saturated solution under specific conditions.
This area of chemistry is crucial because it determines whether a solution can reach equilibrium, and how the substances interact when dissolved.

• For barium nitrate, the solubility is given as 15 g/L; hence, it can be dissolved up to its molar solubility limit in water.
• Beyond this point, any additional barium nitrate will not dissolve, leading to a saturation state.

Understanding solubility is important when discussing saturated solutions and solubility product constants, as these values are used to predict reactions involving ionic compounds in solution.

Chemical Equilibrium

Chemical equilibrium occurs in a chemical reaction when the concentrations of reactants and products remain constant over time, indicating a dynamic balance.
In the context of barium nitrate in water, equilibrium involves the forward and reverse reactions of dissociation and crystallization.

• As barium nitrate dissolves, it reaches a point where its rate of dissolving equals the rate of precipitation or crystallization, maintaining a constant concentration in solution.
• The equilibrium concentrations of the ions are then used to calculate the solubility product constant (\(K_{sp}\)).
• Specifically, \(K_{sp}\) for barium nitrate is determined by the equation: \[K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{NO}_{3}^-]^2 = (S)(2S)^2\]

This balance between the dissolved ions and the solid compound is fundamental in calculating the solubility product constant and understanding solution dynamics.