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Chapter 17: Problem 68

(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.020 \mathrm{M}\) solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is adjusted to \(8.5 ?\) (b) Will \(\mathrm{AgIO}_{3}\) precipitate when \(20 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{NaIO}_{3} ?\left(K_{s p}\right.\) of \(\mathrm{AgIO}_{3}\) is \(\left.3.1 \times 10^{-8} .\right)\)

Short Answer

Expert verified
(a) Co(OH)2 will not precipitate because the calculated ion product (IP) is smaller than the solubility product constant (Ksp) of Co(OH)2. (b) AgIO3 will precipitate since the ion product (IP) is greater than the solubility product constant (Ksp) of AgIO3.

Step by step solution

01

Find the OH- concentration

Given the pH of the solution is 8.5, we can calculate the pOH using the formula: pOH = 14 - pH pOH = 14 - 8.5 pOH = 5.5 Now, let's find the concentration of OH- ions. The formula for calculating the concentration of OH- ions is: \[OH^-] = 10^{-pOH}\] \[OH^-] = 10^{-5.5}\]
02

Find the Ksp of Co(OH)2

We can find the Ksp of Co(OH)2 from a reference table. The solubility product constant Ksp for Co(OH)2 is 1.6 × 10^-15.
03

Calculate the ion product and compare with Ksp

Now, let's calculate the ion product (IP) for Co(OH)2 using the concentrations of Co2+ and OH- ions: \[IP = [Co^{2+}][OH^-]^2\] Since the concentration of Co2+ ions is 0.020 M and we found the concentration of OH- ions as \(10^{-5.5} M\), we can plug in these values: \[IP = (0.020)(10^{-5.5})^2\] Now, compare the ion product (IP) to the solubility product constant (Ksp) for Co(OH)2: If IP > Ksp, then Co(OH)2 will precipitate. If IP < Ksp, then Co(OH)2 will not precipitate. • Part (b):
04

Calculate the final concentrations of Ag+ and IO3- ions

We are given that 20 mL of 0.010 M AgNO3 is mixed with 10 mL of 0.015 M NaIO3. We must find the final concentrations of Ag+ and IO3- ions. First, let's find the moles of each ion: moles of Ag+ = (20 mL)(0.010 mol/L) moles of IO3- = (10 mL)(0.015 mol/L) Next, we need to find the total volume of the mixture: total volume = 20 mL + 10 mL = 30 mL Now, let's find the final concentrations of both ions: \[Ag^+ = \frac{moles\ of\ Ag^+}{total\ volume}\] \[IO3^- = \frac{moles\ of\ IO3^-}{total\ volume}\]
05

Calculate the ion product for AgIO3

Now, let's calculate the ion product (IP) for AgIO3 using the final concentrations of Ag+ and IO3- ions: \[IP = [Ag^+][IO3^-]\]
06

Compare the ion product to Ksp of AgIO3

We are given that the solubility product constant Ksp for AgIO3 is 3.1 × 10^-8. Now, compare the calculated ion product (IP) to the solubility product constant (Ksp) for AgIO3: If IP > Ksp, then AgIO3 will precipitate. If IP < Ksp, then AgIO3 will not precipitate.

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Most popular questions from this chapter

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\) (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{+},\) (d) \(\mathrm{Ag}^{+}\) and \(\mathrm{Mn}^{2+}\). Suggest how each mixture might be separated.

(a) Consider the equilibrium \(\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) .\) Using Le Châtelier's principle, explain the effect of the presence of a salt of \(\mathrm{HB}^{+}\) on the ionization of B. (b) Give an example of a salt that can decrease the ionization of \(\mathrm{NH}_{3}\) in solution.

A \(35.0-\mathrm{mL}\) sample of \(0.150 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is titrated with \(0.150 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of base have been added: (a) \(0 \mathrm{~mL},(\mathbf{b})\) \(17.5 \mathrm{~mL},(\mathrm{c}) 34.5 \mathrm{~mL},(\) d) \(35.0 \mathrm{~mL},\) (e) \(35.5 \mathrm{~mL},\) (f) \(50.0 \mathrm{~mL} .\)

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14}\) (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?(\mathbf{b})\) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what would the initial concentration of \(\mathrm{NaBr}\) need to be in order to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3}\) moles per liter?

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C} ?\)
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