Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 82

The equilibrium constant for the following reaction is \(1.0 \times 10^{23}\). EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt \(\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}\), are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 M\) in \(\mathrm{Cr}^{3+}\) and \(0.050 \mathrm{M}\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00 .\)

Short Answer

Expert verified
At equilibrium, the concentration of \(\mathrm{Cr^{3+}}\) ions in the solution is approximately \(0.0010\,\mathrm{M}\).

Step by step solution

01

Reaction Equation

\[\mathrm{Cr^{3+} + H_{2}EDTA^{2-} \rightleftharpoons Cr(EDTA)}\] We have the equilibrium constant Kc = 1.0 × 10²³ for this reaction. Now, let's write the expression for Kc:
02

Equilibrium Constant Expression

\[K_{c} = \frac{\left[\mathrm{Cr}\left(\mathrm{EDTA}\right)\right]}{\left[\mathrm{Cr^{3+}}\right]\left[\mathrm{H_{2}EDTA^{2-}}\right]}\] Next, we will set up an ICE (Initial, Change, Equilibrium) table for the reaction to find equilibrium concentrations:
03

ICE Table and Equilibrium Concentrations

ICE Table: Initial: \[ [\mathrm{Cr^{3+}}] = 0.0010\,\mathrm{M }\] \[ [\mathrm{H_{2}EDTA^{2-}}] = 0.050\,\mathrm{M} \] \[ [\mathrm{Cr(EDTA)}] = 0\,\mathrm{M} \] Change: \[ [\mathrm{Cr^{3+}}] = -x\,\mathrm{M} \] \[ [\mathrm{H_{2}EDTA^{2-}}] = -x\,\mathrm{M} \] \[ [\mathrm{Cr(EDTA)}] = +x\,\mathrm{M} \] Equilibrium: \[ [\mathrm{Cr^{3+}}] = (0.0010 - x)\,\mathrm{M} \] \[ [\mathrm{H_{2}EDTA^{2-}}] = (0.050 - x)\,\mathrm{M} \] \[ [\mathrm{Cr(EDTA)}] = x\,\mathrm{M} \] Now, we can substitute the equilibrium concentrations into the expression for Kc:
04

Substitution of Equilibrium Concentrations

\[\begin{aligned} 1.0 \times 10^{23} &= \frac{x}{(0.0010 - x)(0.050 - x)} \\ \end{aligned}\] Now, solve the equation for x:
05

Solving for x

Let's assume that the changes (x) are too small compared to the initial concentrations, meaning that the denominator remains essentially constant: \((0.0010)(0.050)\). This simplifies the expression as: \[1.0 \times 10^{23} = \frac{x}{(0.0010)(0.050)}\] Solving for x: \[x = 1.0 \times 10^{23} \times (0.0010)(0.050)\] \[x \approx 5.0 \times 10^{-22}\,\mathrm{M}\] Now that we have the value of x, we can find the equilibrium concentration of Cr^3+:
06

Calculating Equilibrium Concentration of Cr^3+

\[[\mathrm{Cr^{3+}}]_{equilibrium} = 0.0010 - x\] \[[\mathrm{Cr^{3+}}]_{equilibrium} = 0.0010 - 5.0 \times 10^{-22}\] \[[\mathrm{Cr^{3+}}]_{equilibrium} \approx 0.0010\,\mathrm{M}\] So, at equilibrium, the concentration of Cr^3+ ions in the solution is approximately 0.0010 M.

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

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{CoF}_{6}^{3-}\) b. \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\)

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 \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 \mathrm{M}\) \(\mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 \mathrm{M} .\)

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 \(\mathrm{HgI}_{4}{ }^{2-}\).)

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{Ag}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+}\right.\) is the cation in solution.)

Sodium tripolyphosphate \(\left(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\right)\) is used in many synthetic detergents. Its major effect is to soften the water by complexing \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) ions. It also increases the efficiency of surfactants, or wetting agents that lower a liquid's surface tension. The \(K\) value for the formation of \(\mathrm{MgP}_{3} \mathrm{O}_{10}^{3-}\) is \(4.0 \times 10^{8} .\) The reaction is \(\mathrm{Mg}^{2+}+\mathrm{P}_{3} \mathrm{O}_{10}^{5-} \rightleftharpoons \mathrm{MgP}_{3} \mathrm{O}_{10}{ }^{3-} .\) Calculate the concentration of \(\mathrm{Mg}^{2+}\) in a solution that was originally \(50 . \mathrm{ppm} \mathrm{Mg}^{2+}(50 . \mathrm{mg} / \mathrm{L}\) of solution) after 40. g \(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\) is added to \(1.0 \mathrm{~L}\) of the solution.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

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