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Chapter 17: Problem 95

Radiochemical techniques are useful in estimating the solubility product of many compounds. In one experiment, \(50.0 \mathrm{~mL}\) of a \(0.010 \mathrm{M} \mathrm{AgNO}_{3}\) solution containing a silver isotope with a radioactivity of 74,025 counts per min per \(\mathrm{mL}\) were mixed with \(100 \mathrm{~mL}\) of a \(0.030 \mathrm{M} \mathrm{NaIO}_{3}\) solution. The mixed solution was diluted to \(500 \mathrm{~mL}\) and filtered to remove all of the \(\mathrm{AgIO}_{3}\) precipitate. The remaining solution was found to have a radioactivity of 44.4 counts per min per \(\mathrm{mL}\). What is the \(K_{\mathrm{sp}}\) of \(\mathrm{AgIO}_{3} ?\)

Short Answer

Expert verified
The solubility product (Ksp) of AgIO3 is \(7.84 \times 10^{-8}\).

Step by step solution

01

Find initial moles of AgNO3 and NaIO3

Calculate the initial moles of AgNO3 and NaIO3 before mixing. For AgNO3: Volume = 50.0 mL = 0.050 L, Concentration = 0.010 M Initial moles of AgNO3 = 0.050 L x 0.010 mol/L = 0.0005 mol For NaIO3: Volume = 100.0 mL = 0.100 L, Concentration = 0.030 M Initial moles of NaIO3 = 0.100 L x 0.030 mol/L = 0.003 mol.
02

Determine initial and final silver ion concentrations

Calculate initial concentration of silver ions from AgNO3 before dilution: Initial concentration of Ag鈦� = 0.010 M (since 1:1 stoichiometry in AgNO3). Determine radioactive counts for initial silver ions: Initial total radiation = 50 mL x 74,025 counts/min/mL = 3,701,250 counts/min. After the reaction: Total remaining radiation = 500 mL x 44.4 counts/min/mL = 22,200 counts/min. The final concentration of silver ions in the diluted solution after precipitation = 44.4 / 74,025 M = 0.0006 M.
03

Account for dilution and find final concentration

Calculate the dilution factor. Total volume after mixing = 50 mL + 100 mL = 150 mL. Dilution factor to 500 mL = 500 mL / 150 mL = 10/3. Final concentration of Ag鈦� after precipitation and dilution: Initial moles of Ag鈦� remaining = 0.050 L x 0.010 M - 0.0005 mol precipitated = 0.00014 mol remaining Final concentration of remaining silver ions after dilution = (0.00014 mol / 0.500 L) = 0.00028 M.
04

Calculate final concentration of IO3- ions

Assume all AgIO3 precipitated, attempt to find equilibrium concentration of IO3鈦�. NaIO3 initially gives IO3- moles = 0.003 mol. Remaining IO3- after reaction in 500 mL new solution: (0.00028 M remaining Ag+) implies matching IO3- = 0.00028 M at equilibrium.
05

Calculate Ksp for AgIO3

Using the equilibrium concentrations:Ksp = \[[Ag^+][IO_3^-] = (0.00028)\times(0.00028)\]Insert values and solve:Ksp = (0.00028)\times(0.00028) = 7.84 \times 10^{-8}.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radiochemical Techniques
Radiochemical techniques involve using radioactive isotopes to study chemical processes. These methods are valuable in estimating solubility products like that of AgIO鈧�. By tracking the radioactivity, scientists can gain insights into the concentrations of specific ions in a solution.

This method is useful in this experiment with silver isotopes introduced via AgNO鈧�. The initial radioactivity here helps to determine the concentration of silver ions before and after precipitation.
  • Tracking the change in radioactivity allows calculation of ion concentrations.
  • It provides a non-invasive way to study chemical equilibria without altering the sample significantly.
This makes radiochemical techniques a powerful tool in modern chemistry.
AgIO3 Precipitation
AgIO鈧� or Silver Iodate precipitation is an important step in the described experiment.

When solutions of AgNO鈧� and NaIO鈧� are mixed, they undergo a reaction to form AgIO鈧� as a precipitate. This occurs because AgIO鈧� is less soluble in water.
  • The process effectively removes Ag鈦� ions from the solution by forming a solid compound.
  • This allows the measurement of remaining Ag鈦� ions in the solution, which is crucial for calculating the solubility product K鈧涒倸 of the precipitate.
The ability to precipitate out a component is key in determining the concentrations of ions that remain in equilibrium.
Equilibrium Concentration
Equilibrium concentration is the stability point where the concentrations of ions in solution remain constant over time.

After the AgIO鈧� precipitates, the solution reaches a state of equilibrium. Here, the concentration of remaining Ag鈦� ions in solution tells us about the dissolved amount of the ions after precipitation.
  • The process involves calculating the concentration of Ag鈦� and IO鈧冣伝 ions that remain dissolved.
  • These concentrations are essential for deriving the K鈧涒倸 value, which quantifies the solubility of a slightly soluble compound like AgIO鈧�.
In essence, equilibrium concentrations help us understand the solubility and reaction dynamics of chemical processes.
Dilution Factor
The dilution factor reflects how much a solution is diluted by additional solvent, and it is crucial for calculating concentrations in a larger volume.

In this experiment, the initial solution of 150 ml is diluted to 500 ml.
  • This increases the total volume, decreasing the molarity of ions proportionally.
  • Using the formula Dilution Factor = (Total Volume After Dilution) / (Initial Total Volume), we determine the scale of dilution in the experiment.
By applying this factor, we adjust the initial calculated concentrations to reflect the final, smaller concentrations in the solution. This is key to accurately determining the equilibrium concentration and subsequently the K鈧涒倸 value after a dilution process.

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Most popular questions from this chapter

What reagents would you employ to separate these pairs of ions in solution: (a) \(\mathrm{Na}^{+}\) and \(\mathrm{Ba}^{2+}\), (b) \(\mathrm{K}^{+}\) and \(\mathrm{Pb}^{2+},\) (c) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+} ?\)

The molar solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) in a \(0.10 \mathrm{M} \mathrm{NaIO}_{3}\) solution is \(2.4 \times 10^{-11} \mathrm{~mol} / \mathrm{L}\). What is \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2} ?\)

How does a common ion affect solubility? Use Le Ch芒telier's principle to explain the decrease in solubility of \(\mathrm{CaCO}_{3}\) in a \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution.

Calculate the \(\mathrm{pH}\) at the equivalence point for these titrations: (a) \(0.10 M \mathrm{HCl}\) versus \(0.10 \mathrm{M} \mathrm{NH}_{3}\), (b) \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) versus \(0.10 \mathrm{M} \mathrm{NaOH}\).

A volume of \(75 \mathrm{~mL}\) of \(0.060 \mathrm{M} \mathrm{NaF}\) is mixed with \(25 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2} .\) Calculate the concentrations in the final solution of \(\mathrm{NO}_{3}^{-}, \mathrm{Na}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{F}^{-}\). \(\left(K_{\mathrm{sp}}\right.\) for \(\left.\mathrm{SrF}_{2}=2.0 \times 10^{-10} .\right)\)
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