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\(\mathrm{~A} 1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution is slowly added to \(10.0 \mathrm{~mL}\) of a solution that is \(0.20 \mathrm{M}\) in \(\mathrm{Ca}^{2+}\) and \(0.30 \mathrm{M}\) in \(\mathrm{Ag}^{+}\). (a) Which compound will precipitate first: \(\mathrm{CaSO}_{4}\left(K_{s p}=2.4 \times 10^{-5}\right)\) or \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{s p}=1.5 \times 10^{-5}\right) ?(\mathbf{b})\) How much \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) solution must be added to initiate the precipitation?

Short Answer

Expert verified
CaSO4 will precipitate first. 12 mL of the Na2SO4 solution must be added to initiate the precipitation of CaSO4.

Step by step solution

01

Determine the reaction quotient (Q) for both compounds

Calculate the reaction quotient Q for both CaSO4 and Ag2SO4 using their respective concentrations. The formula for Q is given by: Q = [concentration of ions in the product side]^stoichiometric_coefficients For CaSO4 and Ag2SO4, the expressions for Q can be written as: Q(CaSO4) = [Ca虏鈦篯[SO鈧劼测伝] Q(Ag2SO4) = [Ag鈦篯虏[SO鈧劼测伝] Given concentrations: [Ca虏鈦篯 = 0.20 M [Ag鈦篯 = 0.30 M [SO鈧劼测伝] = 1.0 M (from Na2SO4 solution) Calculate Q for both compounds: Q(CaSO4) = (0.20)(1.0) = 0.20 Q(Ag2SO4) = (0.30)虏(1.0) = 0.09
02

Compare Q values to the solubility product constants (Ksp)

Now, let's compare the calculated Q values to their respective solubility product constants (Ksp). If Q > Ksp, the compound will precipitate; if Q < Ksp, the compound will not precipitate. Ksp(CaSO4) = 2.4 脳 10鈦烩伒 Ksp(Ag2SO4) = 1.5 脳 10鈦烩伒 Comparing Q to Ksp: Q(CaSO4) > Ksp(CaSO4): 0.20 > 2.4 脳 10鈦烩伒 (calcium sulfate will precipitate) Q(Ag2SO4) < Ksp(Ag2SO4): 0.09 < 1.5 脳 10鈦烩伒 (silver sulfate will not precipitate) So, CaSO4 will precipitate first.
03

Calculate the volume of Na2SO4 solution required to initiate precipitation

Now we want to calculate the volume of the Na2SO4 solution that is needed to initiate the precipitation of the first compound (CaSO4). We know the Ksp expression for CaSO4: Ksp(CaSO4) = [Ca虏鈦篯[SO鈧劼测伝] We can rearrange this equation to solve for [SO鈧劼测伝]: [SO鈧劼测伝] = Ksp(CaSO4) / [Ca虏鈦篯 Now we know that the concentration of SO鈧劼测伝 required to initiate precipitation is equal to the concentration of SO鈧劼测伝 in the Na2SO4 solution: [SO鈧劼测伝] = V(Na2SO4) * [Na2SO4] / (V(Na2SO4) + V(Ca鈦衡伜 and Ag鈦 solution)) Now, plug in all the given values and solve for V(Na2SO4): Ksp(CaSO4) / [Ca虏鈦篯 = (V(Na2SO4) * [Na2SO4]) / (V(Na2SO4) + V(Ca鈦衡伜 and Ag鈦 solution)) (2.4 脳 10鈦烩伒) / 0.20 = (V(Na2SO4) * 1.0) / (V(Na2SO4) + 10.0) Now solve for V(Na2SO4): V(Na2SO4) = 12 mL So, 12 mL of the Na2SO4 solution must be added to initiate the precipitation of CaSO4.

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Key Concepts

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

Reaction Quotient (Q)
Understanding the reaction quotient (Q) is essential for predicting whether a precipitate will form in a chemical reaction. The reaction quotient is often compared to the solubility product constant (Ksp) to determine the direction of a precipitation reaction.

To calculate Q, you take the product of the concentrations of the ions involved in the reaction, each raised to the power of their stoichiometric coefficients found in the balanced equation for the dissolution of the solid. This calculation is crucial for working out whether a precipitate will form under certain conditions. For example, if Q is greater than Ksp, it indicates that the system is supersaturated and a precipitate will form. In contrast, if Q is less than Ksp, the system is unsaturated and precipitation won't occur. Understanding Q allows you to predict and explain the behavior of ions in solution, particularly in the context of solubility and precipitation reactions.
Solubility Product Constant (Ksp)
The solubility product constant (Ksp) is a crucial concept in chemistry, particularly when discussing the solubility of ionic compounds. This value represents the extent to which a compound can dissolve in water and is unique for each solute at a given temperature.

The Ksp expression for a salt, which is the product of the ion concentrations raised to the power of their stoichiometric coefficients, allows us to understand the point at which a solution becomes saturated and precipitation begins. For instance, in our exercise, the Ksp values for calcium sulfate (CaSO_4) and silver sulfate (Ag_2SO_4) are compared to the Q values to deduce which compound will precipitate first. Being familiar with how to use the Ksp value can allow students to make accurate predictions about the solubility of different compounds in chemistry problems.
Precipitation Calculations
Precipitation calculations are a vital aspect of chemistry, particularly in the field of analytical chemistry, where quantitative analysis and understanding the composition of solutions are key.

To determine when a precipitate will form, as shown in our exercise solution, you can perform calculations involving the reaction quotient and the solubility product constant. These calculations enable us to predict the quantity of a solution needed to initiate precipitation. This process requires a careful balance between the ions in the solution and the stoichiometry of the precipitate. This knowledge is not only important for laboratory chemists but also in industrial processes where controlling precipitation is often critical to the production or treatment of chemicals.
Stoichiometry
Stoichiometry is the portion of chemistry that deals with the relationships between the amounts of reactants and products in a chemical reaction. It provides the quantitative basis for the conservation of mass, often referred to as 'the recipe' of a reaction.

In understanding precipitation reactions, stoichiometry allows us to calculate the exact amount of reactants needed to produce a given amount of product, and likewise, predict the amount of product formed from given reactants. An understanding of stoichiometry is indispensable when performing precipitation calculations, as it ensures that the conservation of mass is observed and that predictions made are accurate. This core concept is deeply intertwined with the exercises concerning solubility and precipitation.

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Most popular questions from this chapter

Assume that \(30.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of a weak base \(\mathrm{B}\) that accepts one proton is titrated with a \(0.10 \mathrm{M}\) solution of the monoprotic strong acid HX. (a) How many moles of \(\mathrm{HX}\) have been added at the equivalence point? (b) What is the predominant form of \(\mathrm{B}\) at the equivalence point? (c) What factor determines the \(\mathrm{pH}\) at the equivalence point? (d) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

How many milliliters of \(0.0850 \mathrm{M} \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) \(40.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3}\), (b) \(35.0 \mathrm{~mL}\) of \(0.0850 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH},\) (c) \(50.0 \mathrm{~mL}\) of a solution that contains \(1.85 \mathrm{~g}\) of HCl per liter?

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\) (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{+},\) (d) \(\mathrm{Ag}^{+}\) and \(\mathrm{Mn}^{2+}\). Suggest how each mixture might be separated.

A sample of \(0.2140 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M} \mathrm{NaOH}\). The acid required \(27.4 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be 6.50 . What is the \(K_{a}\) for the unknown acid?

Using the value of \(K_{s p}\) for \(\mathrm{Ag}_{2} \mathrm{~S}, K_{a 1}\) and \(K_{a 2}\) for \(\mathrm{H}_{2} \mathrm{~S},\) and \(K_{f}=1.1 \times 10^{5}\) for \(\mathrm{AgCl}_{2}^{-},\) calculate the equilibrium constant for the following reaction: \(\mathrm{Ag}_{2} \mathrm{~S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{~S}(a q)\)

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