Question

\mathrm{~A}\( solution of \)\mathrm{Na}_{2} \mathrm{SO}_{4}\( is added dropwise to a solution that is \)0.010 \mathrm{M}_{\text {in } \mathrm{Ba}^{2+}}\( and \)0.010 \mathrm{M}\( in \)\mathrm{Sr}^{2+}\(. (a) What concentration of \)\mathrm{SO}_{4}^{2-}\( is necessary to begin precipitation? (Neglect volume changes, \)\mathrm{BaSO}_{4} ; K_{p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}\( \)K_{s p}=3.2 \times 10^{-7}\( ) (b) Which cation precipitates first? (c) What is the concentration of \)\mathrm{SO}_{4}^{2-}$ when the second cation begins to precipitate?

Short answer

The concentration of SO₄²⁻ necessary to begin precipitation is \(1.1 \times 10^{-8}\) M. Ba²⁺ cation precipitates first. The concentration of SO₄²⁻ when the second cation, Sr²⁺, begins to precipitate is \(3.2 \times 10^{-5}\) M.

Step-by-step solution

Find the saturation concentration of SO₄²⁻ for Ba²⁺ and Sr²⁺

To determine the saturation concentration of SO₄²⁻ for each cation, we will use Ksp values and the given concentration of each cation. For BaSO₄: \(K_{sp} = [Ba^{2+}] [SO_4^{2-}]\) Given \(K_{sp} = 1.1 \times 10^{-10}\) and \([Ba^{2+}] = 0.010 M\), we can solve for \([SO_4^{2-}]\). For SrSO₄: \(K_{sp} = [Sr^{2+}] [SO_4^{2-}]\) Given \(K_{sp} = 3.2 \times 10^{-7}\) and \([Sr^{2+}] = 0.010 M\), we can solve for \([SO_4^{2-}]\).

Calculate the concentration of SO₄²⁻ required for each cation

For BaSO₄, using the formula \(K_{sp} = [Ba^{2+}] [SO_4^{2-}]\), we have: \(1.1 \times 10^{-10} = (0.010) [SO_4^{2-}]\) \([SO_4^{2-}] = \frac{1.1 \times 10^{-10}}{0.010}\) \([SO_4^{2-}] = 1.1 \times 10^{-8}\) For SrSO₄, using the formula \(K_{sp} = [Sr^{2+}] [SO_4^{2-}]\), we have: \(3.2 \times 10^{-7} = (0.010) [SO_4^{2-}]\) \([SO_4^{2-}] = \frac{3.2 \times 10^{-7}}{0.010}\) \([SO_4^{2-}] = 3.2 \times 10^{-5}\)

Determine which cation precipitates first

Comparing the saturation concentrations of SO₄²⁻ calculated in Step 2, we can see that Ba²⁺ requires a lower concentration of SO₄²⁻ (\(1.1 \times 10^{-8}\) M) to precipitate compared to Sr²⁺ (\(3.2 \times 10^{-5}\) M). Therefore, Ba²⁺ will precipitate first.

Calculate the concentration of SO₄²⁻ when the second cation begins to precipitate

Since Ba²⁺ is the first cation to precipitate, we now consider the concentration of Sr²⁺ in the remaining solution (since all Ba²⁺ has precipitated). We have to use the solubility product for SrSO₄ to find the concentration of SO₄²⁻ needed for Sr²⁺ to start precipitating. We have: \(3.2 \times 10^{-7} = [Sr^{2+}] [SO_4^{2-}]\) Since the concentration of Sr²⁺ remains the same (\([Sr^{2+}] = 0.010 M\)), we can find the concentration of SO₄²⁻: \([SO_4^{2-}] = \frac{3.2 \times 10^{-7}}{0.010}\) \([SO_4^{2-}] = 3.2 \times 10^{-5} M\) The concentration of SO₄²⁻ needed for the second cation (Sr²⁺) to begin precipitating is \(3.2 \times 10^{-5}\) M.

Key concepts

Ksp Calculations

Understanding the solubility product constant (Ksp) is essential in predicting the solubility of sparingly soluble salts in water. The solubility product is a special kind of equilibrium constant that measures the extent of dissolution for ionic compounds.

Let's demystify how Ksp calculations are done, which involve setting up an equilibrium expression for the dissociation of a salt. Consider the generic salt AB, having the equation AB(s) ↔ A+(aq) + B-(aq). The Ksp would be represented as:\[ K_{sp} = [A^+][B^-] \]In this formula, [A+] and [B-] represent the molar concentrations of the ions at the point of saturated solution. To find the molar solubility (the concentration of a saturated solution), we rearrange the expression for Ksp to solve for either ion's molar concentration.

Applying this to the example in our exercise involving barium sulfate, the equation would look like this:\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \]Given the Ksp and the concentration of Ba²⁺, we calculated the concentration of sulfate ions needed to reach the saturation point. This calculation helps us to know how much solute can be dissolved before precipitation starts, thus advising on solubility limits of compounds in chemical solutions.

Precipitation Reactions

A precipitation reaction is an intriguing chemical process where ions in aqueous solution combine to form an insoluble solid, called a precipitate. These reactions are crucial in fields ranging from industrial chemistry to environmental science and medicine.

To predict when a precipitate will form, we consider the Ksp and the ion concentrations in solution. The moment the product of these concentrations exceeds the Ksp value, a precipitate begins to form. This is because the solution becomes supersaturated, and excess ions start to aggregate into solid form.

Let's reflect on the context of our textbook example: when sodium sulfate is added to the solution containing Ba²⁺ and Sr²⁺ ions. As the sulfate ions increase in concentration, they eventually reach a level where the solubility product for either barium sulfate or strontium sulfate is exceeded. This marks the point of precipitation, starting with the salt that has the lower Ksp value, hence less soluble.

Cation Precipitation Order

Determining the order in which cations will precipitate from a solution containing multiple cations is a key process in analytical and preparative chemistry. To determine the cation precipitation order, we look at the Ksp values of possible precipitates and the initial concentrations of the cations involved.

The cation with the lowest product of its concentration and the corresponding anion concentration will precipitate first. This is due to its lower solubility reflected in a smaller Ksp value. In the given exercise, when comparing the Ksp values and the concentrations of Ba²⁺ and Sr²⁺ with sulfate ions, we found that barium sulfate has the lower Ksp. Therefore, Ba²⁺ will precipitate before Sr²⁺.

Understanding and predicting the sequence of cation precipitation is fundamental in procedures such as fractional precipitation and qualitative analysis. It helps in separating and identifying different ions present in a solution by gradually adjusting conditions to precipitate out specific cations one at a time.