Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 52

The concentration of Mg \(^{2+}\) in seawater is 0.052\(M .\) At what pH will 99\(\%\) of the \(\mathrm{Mg}^{2+}\) be precipitated as the hydroxide salt? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Mg}(\mathrm{OH})_{2}=8.9 \times 10^{-12} .\right]$

Short Answer

Expert verified
The pH at which 99% of the Mg2+ will be precipitated as the hydroxide salt is approximately 10.12.

Step by step solution

01

Calculate the Concentration of Mg2+ Ions After 99% Precipitation

To find the concentration of Mg2+ ions after 99% precipitation, we use the original concentration and the percentage given: \(0.99 \times 0.052 M = 0.05148 M\) So, the concentration of Mg2+ ions that would remain in the solution is: \(0.052 M - 0.05148 M = 0.00052 M\)
02

Set Up the Ion Product Expression for Mg(OH)2

The ion product expression (Q) for Mg(OH)2 is given by \(Q = [Mg^{2+}][OH^-]^2\) Now, we know that at the point of precipitation, Q = Ksp. Therefore \(K_{sp} = [Mg^{2+}][OH^-]^2\) We know the Ksp value for Mg(OH)2 is 8.9 x 10^(-12) and the concentration of Mg2+ left in the solution is 0.00052 M. So, \(8.9 \times 10^{-12} = [0.00052 M][OH^-]^2\)
03

Solve for the Concentration of OH- Ions

Now, we can solve for the concentration of OH- ions. \([OH^-]^2 = \frac{8.9 \times 10^{-12}}{0.00052 M}\) \([OH^-]^2 = 1.71 \times 10^{-8}\) Taking the square root of both sides, we get: \([OH^-] \approx 1.31 \times 10^{-4}\)
04

Calculate the pH Using the Concentration of OH- Ions

We can now use the concentration of OH- ions to calculate the pH of the solution. First, we need to find the pOH. \(pOH = -\log_{10}(OH^-)\) \(pOH = -\log_{10}(1.31 \times 10^{-4})\) \(pOH \approx 3.88\) Now, we can find the pH using the relation, \(pH + pOH = 14\) Thus, \(pH = 14 - pOH\) \(pH = 14 - 3.88\) \(pH \approx 10.12\) #Conclusion#: The pH at which 99% of the Mg2+ will be precipitated as the hydroxide salt is approximately 10.12.

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

When aqueous KI is added gradually to mercury (II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{HgI}_{4}^{2-}.)\)

The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a $0.20-M \mathrm{KIO}_{3}\( solution is \)4.4 \times 10^{-8} \mathrm{mol} / \mathrm{L}$ . Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}.\)

Magnesium hydroxide, \(\operatorname{Mg}(\mathrm{OH})_{2},\) is the active ingredient in the antacid TUMS and has a \(K_{\mathrm{sp}}\) value of $8.9 \times 10^{-12}\( . If a 10.0 -g sample of \)\mathrm{Mg}(\mathrm{OH})_{2}$ is placed in \(500.0 \mathrm{mL}\) of solution, calculate the moles of OH -ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is much less than \(1,\) not a lot solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acid.

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 \(0.0095 M\) .

A solution is formed by mixing \(50.0 \mathrm{mL}\) of \(10.0 \mathrm{M}\) \(\mathrm{NaX}\) with \(50.0 \mathrm{mL}\) of $2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3} .\( Assume that \)\mathrm{Cu}^{+}$ forms complex ions with \(\mathrm{X}^{-}\) as follows: $$\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) \quad K_{1}=1.0 \times 10^{2}$$ $$\operatorname{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) \qquad K_{2}=1.0 \times 10^{4}$$ $$\operatorname{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K_{3}=1.0 \times 10^{3}$$ with an overall reaction $$\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \qquad K=1.0 \times 10^{9}$$ Calculate the following concentrations at equilibrium a. \(\mathrm{CuX}_{3}^{2-} \quad\) b. \(\mathrm{CuX}_{2}^{-} \quad\) c. \(\mathrm{Cu}^{+}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Organic 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