/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 A solution is \(1 \times 10^{-4}... [FREE SOLUTION] | 91影视

91影视

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}, \mathrm{Na}_{2} \mathrm{S},\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are \(K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)\) \(=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},\) and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=\) \(1 \times 10^{-54}\).

Short Answer

Expert verified
The order of precipitation as a source of Pb虏鈦 is added gradually to the solution is: PbF鈧 鈫 PbS 鈫 Pb鈧(PO鈧)鈧.

Step by step solution

01

Write the solubility equilibrium expressions for each salt

The solubility equilibrium expressions for PbF鈧, PbS, and Pb鈧(PO鈧)鈧 based on their Ksp values: 1. PbF鈧 鈬 Pb虏鈦 + 2F鈦 2. PbS 鈬 Pb虏鈦 + S虏鈦 3. Pb鈧(PO鈧)鈧 鈬 3Pb虏鈦 + 2PO鈧劼斥伝
02

Calculate the concentration of Pb虏鈦 required to reach Ksp for each salt

To find the concentration of Pb虏鈦 required for each salt to reach its Ksp value, set up equations using the solubility equilibrium expressions and solve for the concentration of Pb虏鈦. 1. For PbF鈧: Ksp(PbF鈧) = [Pb虏鈦篯[F鈦籡虏 4脳10鈦烩伕 = (x)(2脳10鈦烩伌)虏 Solve for x to find the concentration of Pb虏鈦 required to reach Ksp for PbF鈧. 2. For PbS: Ksp(PbS) = [Pb虏鈦篯[S虏鈦籡 7脳10鈦宦测伖 = (x)(1脳10鈦烩伌) Solve for x to find the concentration of Pb虏鈦 required to reach Ksp for PbS. 3. For Pb鈧(PO鈧)鈧: Ksp(Pb鈧(PO鈧)鈧) = [Pb虏鈦篯鲁[PO鈧劼斥伝]虏 1脳10鈦烩伒鈦 = (x)鲁(1脳10鈦烩伌)虏 Solve for x to find the concentration of Pb虏鈦 required to reach Ksp for Pb鈧(PO鈧)鈧.
03

Determine the order of precipitation

After solving for the concentration of Pb虏鈦 required to reach Ksp for each salt, compare the values to determine the order of precipitation. The salt with the lowest required concentration of Pb虏鈦 will precipitate first, followed by the salt with the next lowest concentration, and so on. PbF鈧 requires less Pb虏鈦 to reach its Ksp value compared to PbS and Pb鈧(PO鈧)鈧. So, PbF鈧 will precipitate first. Next, compare the values for PbS and Pb鈧(PO鈧)鈧. PbS requires less Pb虏鈦 to reach its Ksp value compared to Pb鈧(PO鈧)鈧. Therefore, PbS will precipitate second, and Pb鈧(PO鈧)鈧 will precipitate last. The order of precipitation is: PbF鈧 鈫 PbS 鈫 Pb鈧(PO鈧)鈧.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Ksp values
Understanding the solubility product constant, commonly referred to as Ksp, is crucial when delving into the solubility of ionic compounds in aqueous solutions. The Ksp value represents the extent to which a compound can dissolve in water and is a measure of the product of the molar concentrations of the ions in a saturated solution, each raised to the power of its coefficient in the balanced equation.

These values allow chemists to determine whether a precipitate will form when two solutions are mixed. A precipitate is a solid that forms when the product of the ion concentrations exceeds the Ksp. If the product of the concentrations (called the reaction quotient, Q) is lower than the Ksp, no precipitation will occur and the ions will remain dissolved. When the Q equals the Ksp, the solution is at equilibrium, and the rate at which the solid dissolves is equal to the rate at which it precipitates.

In the described exercise, knowledge of Ksp values for PbF鈧 (\(4 \times 10^{-8}\)), PbS (\(7 \times 10^{-29}\)), and Pb鈧(PO鈧)鈧 (\(1 \times 10^{-54}\)) is used to foresee the order of precipitation. The lower the Ksp, the less soluble the compound is, and consequently, the more likely it is to precipitate out of solution at lower concentrations of the ions.
Precipitation Reaction
The concept of a precipitation reaction revolves around the formation of an insoluble solid from the reaction of two soluble compounds in aqueous solution. Precipitation reactions are a subset of chemical reactions, where ionic compounds exchange ions and produce at least one insoluble product, the precipitate.

Using the rules of solubility and the solubility product constant (Ksp), it's possible to predict which compound would precipitate first in a given situation. One example is adding a source of Pb虏鈦 ions to a solution containing fluoride, sulfate, and phosphate ions. The order of precipitation will depend on the relative solubility of the potential precipitates that can form, which is directly linked to the Ksp values of their corresponding salts.

The exercise demonstrates this by showing that as Pb虏鈦 is gradually added, the solid that forms first is the one associated with the highest solubility product (lowest Ksp), since it needs a smaller concentration of Pb虏鈦 to exceed its solubility threshold and begin to precipitate.
Concentration Calculation
The process of concentration calculation is fundamental in predicting the occurrence of precipitation reactions. It involves calculating the molarity, or molar concentration, of ions in solution, which is crucial when comparing to Ksp values to assess saturation levels and the potential for precipitate formation.

In the context of the exercise provided, the calculation required solving for the concentration of Pb虏鈦 ions that would lead to the saturated solution conditions for each salt. This involves algebraically manipulating the solubility equilibrium expressions and the provided initial ion concentrations. Once computed, these concentrations reveal the threshold at which adding more Pb虏鈦 ions will result in exceeding the Ksp value and precipitating a corresponding salt.

For example, to find the concentration of Pb虏鈦 needed for PbF鈧 to reach its Ksp, we set up the equation \(4 \times 10^{-8} = (x)(2 \times 10^{-4})^2\) and solve for x. The calculated concentrations allow us to compare the solubility of each lead salt and determine the order in which they will precipitate from the solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(100.0 \mathrm{mL}\) of \(1.0 \times 10^{-2}\) \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{mL}\) of \(1.0 \times 10^{-3} \mathrm{M}\) NaF. Will \(\mathrm{PbF}_{2}(s)\) \(\left(K_{\mathrm{sp}}=4 \times 10^{-8}\right)\) precipitate?

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are \(8.5 \times 10^{-40} M\) and \(1.5 \times 10^{-3} M,\) respectively, in a \(0.11-M\) KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \quad K_{\mathrm{overall}}=?$$

a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\) calculate the value for the equilibrium constant for the following reaction: $$\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$$ b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\).

A solution contains \(0.25\) \(M\) \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.25 M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) Can the metal ions be separated by slowly adding \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) Assume that for successful separation \(99 \%\) of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\).

A solution is prepared by mixing \(75.0 \mathrm{mL}\) of \(0.020\) \(M\) \(\mathrm{BaCl}_{2}\) and \(125 \mathrm{mL}\) of \(0.040\) \(M\) \(\mathrm{K}_{2} \mathrm{SO}_{4}\). What are the concentrations of barium and sulfate ions in this solution? Assume only \(\mathrm{SO}_{4}^{2-}\) ions \(\left(\text { no } \mathrm{HSO}_{4}^{-}\right)\) are present.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.