Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 59

A 50.0 -mL sample of \(0.00200 M\) \(\mathrm{AgNO}_{3}\) is added to 50.0 \(\mathrm{mL}\) of 0.0100 \(M\) \(\mathrm{NaIO}_{3} .\) What is the equilibrium concentration of \(\mathrm{Ag}^{+}\) in solution? $\left(K_{\mathrm{sp}} \text { for } \mathrm{AgIO}_{3} \text { is } 3.2 \times 10^{-8} .\right)$

Short Answer

Expert verified
The equilibrium concentration of Ag+ in the solution is approximately 5.66 × 10⁻⁵ M.

Step by step solution

01

1. Write the balanced chemical equation for the reaction.

Write the balanced chemical equation between AgNO₃ and NaIO₃: \[ \mathrm{AgNO_3}(aq) + \mathrm{NaIO_3}(aq) \rightarrow \mathrm{AgIO_3}(s) + \mathrm{NaNO_3}(aq)\]
02

2. Calculate the initial concentrations of AgNO₃ and NaIO₃.

Given the volume and molarity of each solution, calculate the moles of each reactant. Moles of AgNO₃ = volume × molarity = (50.0 mL × 0.00200 M) = 0.0500L × \(0.00200 \frac{mol}{L}\) = 0.0001 mol Moles of NaIO₃ = volume × molarity = (50.0 mL × 0.0100 M) = 0.0500L × \(0.0100 \frac{mol}{L}\) = 0.0005 mol
03

3. Determine the limiting reactant and the amount of AgIO₃ that can form.

Calculate the mole ratio of the reactants in the balanced equation, and determine which reactant will run out first. In this case, since AgNO₃ and NaIO₃ react in a 1:1 ratio, and there is more NaIO₃ than AgNO₃, AgNO₃ will be the limiting reactant. From the balanced equation, 1 mol of AgNO₃ produces 1 mol of AgIO₃, so the amount of AgIO₃ that can form is equal to the moles of the limiting reactant, AgNO₃. Moles of AgIO₃ formed = 0.0001 mol (from AgNO₃)
04

4. Write the equilibrium expression for Ksp of AgIO₃.

From the given Ksp value, write the equilibrium expression for the dissolution of AgIO₃. Ksp = \([\mathrm{Ag^+}][\mathrm{IO_3^-}]\)
05

5. Calculate the equilibrium concentrations using Ksp.

Since all the AgNO₃ reacted, the initial moles of Ag+ are completely converted to AgIO₃, leaving no initial Ag+. However, some AgIO₃ will dissolve to establish an equilibrium: \(K_{sp} = 3.2 \times 10^{-8} = [\mathrm{Ag^+}][\mathrm{IO_3^-}]\) Since the concentration of Ag+ and IO₃⁻ are equal at equilibrium, we can simplify the expression. Let x = equilibrium concentration of Ag+ (and IO₃⁻). \(3.2 \times 10^{-8} = x^2\)
06

6. Solve the equation for the equilibrium concentration of Ag+.

Solve the equation for x to find the equilibrium concentration of Ag+. From the previous step, square root both sides of the equation to solve for x: \(x = \sqrt{3.2 \times 10^{-8}}\) After calculating, the equilibrium concentration of Ag+ is: x ≈ 5.66 × 10⁻⁵ M The equilibrium concentration of Ag+ in the solution is approximately 5.66 × 10⁻⁵ 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

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} \mathrm{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 \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$

a. Calculate the molar solubility of \(\mathrm{AgBr}\) in pure water. \(K_{\mathrm{sp}}\) for \(\mathrm{AgBr}\) is \(5.0 \times 10^{-13}\) . b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of \(\mathrm{AgBr}\) will dissolve in \(250.0 \mathrm{mL}\) of $3.0 M\( \)\mathrm{NH}_{3}?$ e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes

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.

If \(10.0 \mathrm{mL}\) of $2.0 \times 10^{-3} M \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\( is added to 10.0 \)\mathrm{mL}$ of a \(\mathrm{pH}=10.0 \mathrm{NaOH}\) solution, will a precipitate form?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

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