Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 55

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

Short Answer

Expert verified
In the case of mixing 100.0 mL of \(1.0 \times 10^{-2} M\) Pb(NO₃)₂ and 100.0 mL of \(1.0 \times 10^{-3} M\) NaF, we find the concentrations of Pb²⁺ and F⁻ ions after mixing to be \(5.0 \times 10^{-3} M\) and \(5.0 \times 10^{-4} M\), respectively. Then, we calculate the ion product Q as \(Q = (5.0 \times 10^{-3})(5.0 \times 10^{-4})^2 = 12.5 \times 10^{-11}\). Comparing Q (\(12.5 \times 10^{-11}\)) to the solubility product constant Ksp of PbF₂ (\(4 \times 10^{-8}\)), we find that Q > Ksp, meaning a precipitate of PbF₂ will form.

Step by step solution

01

Calculate the moles of each ion in the initial solutions

First, let's find the moles of lead (II) ions and fluoride ions in the initial solutions: Moles of Pb²⁺ ions: \( (100.0\,mL) * (1.0 \times 10^{-2}\,M) \) Moles of F⁻ ions: \( (100.0\,mL) * (1.0 \times 10^{-3}\,M) \) (Keep in mind that the volume should be converted to liters when calculating moles.)
02

Calculate the concentrations of the ions after mixing

After the solutions are mixed, the total volume of the solution becomes 200.0 mL. Now, we can calculate the new concentration of each ion as follows: Concentration of Pb²⁺ ions: \( \frac{(100.0\,mL) * (1.0 \times 10^{-2}\,M)}{(200.0\,mL)} \) Concentration of F⁻ ions: \( \frac{(100.0\,mL) * (1.0 \times 10^{-3}\,M)}{(200.0\,mL)} \)
03

Calculate the ion product Q

Now we can find the ion product (Q) of the solution by multiplying the concentrations of the ions: \( Q = [\mathrm{Pb}^{2+}] [\mathrm{F^-}]^2 \) Make sure to square the concentration of F⁻ ions.
04

Compare Q to Ksp

Finally, compare the ion product Q calculated in the previous step to the solubility product constant (Ksp) of PbF2: - If Q > Ksp, a precipitate will form. - If Q < Ksp, there will be no precipitate. - If Q = Ksp, the solution is saturated and on the edge of precipitation. Depending on the result, it will be determined if PbF2 will precipitate or not.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Calculate the equilibrium concentrations of $\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\( \)\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+},$ and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) in a solution prepared by mixing \(500.0 \mathrm{mL}\) of \(3.00M\) \(\mathrm{NH}_{3}\) with $500.0 \mathrm{mL}\( of \)2.00 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ . The step wise equilibria are $$\mathrm{Cu}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{CuNH}_{3}^{2+}(a q) \quad K_{1}=1.86 \times 10^{4}$$ $$\mathrm{CuNH}_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q) \quad K_{2}=3.88 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{2}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{3}^{2+}(a q) \quad K_{3}=1.00 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K_{4}=1.55 \times 10^{2}$$

Sulfide precipitates are generally grouped as sulfides insoluble in acidic solution and sulfides insoluble in basic solution. Explain why there is a difference between the two groups of sulfide precipitates.

What happens to the \(K_{\mathrm{sp}}\) value of a solid as the temperature of the solution changes? Consider both increasing and decreasing temperatures, and explain your answer.

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 \(5.0M\) \(\mathrm{NH}_{3} .\) In \(5.0M\) \(\mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\) .

Silver cyanide \((\mathrm{AgCN})\) is an insoluble salt with \(K_{\mathrm{sp}}=2.2 \times 10^{-12}\) . Compare the effects on the solubility of silver cyanide by addition of \(\mathrm{HNO}_{3}(a q)\) or by addition of \(\mathrm{NH}_{3}(a q).\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation
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.

Sign-up for free