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\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

Expert verified
The concentration of SO鈧劼测伝 necessary to begin precipitation is \(1.1 \times 10^{-8}\) M. Ba虏鈦 cation precipitates first. The concentration of SO鈧劼测伝 when the second cation, Sr虏鈦, begins to precipitate is \(3.2 \times 10^{-5}\) M.

Step by step solution

01

Find the saturation concentration of SO鈧劼测伝 for Ba虏鈦 and Sr虏鈦

To determine the saturation concentration of SO鈧劼测伝 for each cation, we will use Ksp values and the given concentration of each cation. For BaSO鈧: \(K_{sp} = [Ba^{2+}] [SO_4^{2-}]\) Given \(K_{sp} = 1.1 \times 10^{-10}\) and \([Ba^{2+}] = 0.010 M\), we can solve for \([SO_4^{2-}]\). For SrSO鈧: \(K_{sp} = [Sr^{2+}] [SO_4^{2-}]\) Given \(K_{sp} = 3.2 \times 10^{-7}\) and \([Sr^{2+}] = 0.010 M\), we can solve for \([SO_4^{2-}]\).
02

Calculate the concentration of SO鈧劼测伝 required for each cation

For BaSO鈧, 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 SrSO鈧, using the formula \(K_{sp} = [Sr^{2+}] [SO_4^{2-}]\), we have: \(3.2 \times 10^{-7} = (0.010) [SO_4^{2-}]\) \([SO_4^{2-}] = \frac{3.2 \times 10^{-7}}{0.010}\) \([SO_4^{2-}] = 3.2 \times 10^{-5}\)
03

Determine which cation precipitates first

Comparing the saturation concentrations of SO鈧劼测伝 calculated in Step 2, we can see that Ba虏鈦 requires a lower concentration of SO鈧劼测伝 (\(1.1 \times 10^{-8}\) M) to precipitate compared to Sr虏鈦 (\(3.2 \times 10^{-5}\) M). Therefore, Ba虏鈦 will precipitate first.
04

Calculate the concentration of SO鈧劼测伝 when the second cation begins to precipitate

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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) 鈫 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 Ba虏鈦, 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 Ba虏鈦 and Sr虏鈦 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 Ba虏鈦 and Sr虏鈦 with sulfate ions, we found that barium sulfate has the lower Ksp. Therefore, Ba虏鈦 will precipitate before Sr虏鈦.

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.

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Most popular questions from this chapter

From the value of \(K_{f}\) listed in Table \(17.1,\) calculate the concentration of \(\mathrm{Ni}^{2}(a q)\) and \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) that are present at equilibrium after dissolving 1.25 \(\mathrm{g} \mathrm{NiCl}_{2}\) in 100.0 \(\mathrm{mL}\) of 0.20 \(\mathrm{MN} \mathrm{H}_{3}(a q) .\)

What is the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of water saturated with \(\mathrm{CO}_{2}\) at a partial pressure of \(1.10 \mathrm{~atm}\) ? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-2} \mathrm{~mol} / \mathrm{L}-\mathrm{atm}\).

Calculate the solubility of \(\mathrm{Mg}_{\mathrm{g}}(\mathrm{OH})_{2}\) in \(0.50 \mathrm{M} \mathrm{NH} \mathrm{NCl}_{4}\).

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C}\) ?

Furoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of (a) a solution formed by adding \(25.0 \mathrm{~g}\) of furoic acid and \(30.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.250 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(20.0 \mathrm{~mL}\). of \(0.22 \mathrm{M} \mathrm{NaC} \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\) and diluting the total volume to \(125 \mathrm{~mL}\). (c) a solution prepared by adding \(50.0 \mathrm{~mL}\) of \(1.65 \mathrm{M} \mathrm{NaOH}\) solution to \(0.500 \mathrm{~L}\) of \(0.0850 \mathrm{M} \mathrm{HC} \mathrm{H}_{3} \mathrm{O}_{3}\).

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