/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 93 To determine the \(K_{\mathrm{sp... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

To determine the \(K_{\mathrm{sp}}\) value of \(\mathrm{Hg}_{2} \mathrm{I}_{2},\) a chemist obtained a solid sample of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) in which some of the iodine is present as radioactive 131 \(\mathrm{I}\) . The count rate of the \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) sample is \(5.0 \times 10^{11}\) counts per minute per mole of I. An excess amount of \(\mathrm{Hg}_{2} \mathrm{I}_{2}(s)\) is placed into some water, and the solid is allowed to come to equilibrium with its respective ions. A \(150.0-\mathrm{mL}\) sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute. From this information, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) $$ \mathrm{Hg}_{2} \mathrm{I}_{2}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \qquad K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2} $$

Short Answer

Expert verified
The Ksp value for Hg2I2 is \(4.3\times10^{-29}\).

Step by step solution

01

We are given the radioactivity count rate of the solid Hg2I2 sample (5.0 x 10^11 counts per minute per mole of I), and after reaching equilibrium, the radioactivity was measured at 33 counts per minute in the 150 mL saturated solution. To find the moles of 131I in the 150 mL saturated solution, we use the ratio between the initial radioactivity count rate and the radioactivity count rate in the saturated solution: \[ \begin{aligned} \text{Moles of 131I in saturated solution} &= \frac{33 \ \text{counts/min}}{5.0\times10^{11}\ \text{counts/min/mol}}\\ &= 6.6\times10^{-11} \text{moles} \end{aligned} \] #Step 2: Determine the concentration of I- ions in the saturated solution# Now that we have the moles of 131I in the 150 mL saturated solution, we can calculate the concentration of I- ions in the saturated solution:

The volume of saturated solution is 150 mL, which is equal to 0.150 L. So, the concentration of I- ions in the solution can be calculated as: \[ \begin{aligned} \text{Concentration of I- ions} &= \frac{\text{Moles of 131I}}{\text{Volume of saturated solution}}\\ &= \frac{6.6\times10^{-11}\ \text{moles}}{0.150\ \text{L}} \\ &= 4.4\times10^{-10}\ \text{M} \end{aligned} \] #Step 3: Determine the concentration of Hg22+ ions in the solution#
02

According to the balanced chemical equation, 1 mole of Hg2I2 produces 1 mole of Hg22+ and 2 moles of I- ions. We know the concentration of I- ions. Hence, we can find the concentration of Hg22+ ions: \[ \text{Concentration of Hg}_{2}^{2+}\ \text{ions} = \frac{1}{2} \times \text{Concentration of I- ions} = \frac{1}{2} \times 4.4\times10^{-10}\ \text{M} = 2.2\times10^{-10}\ \text{M} \] #Step 4: Calculate the Ksp value for Hg2I2#

Now that we have the concentrations of Hg22+ and I- ions in the saturated solution, we can calculate the Ksp value: \[ \begin{aligned} K_{sp} &= \left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}\\ &= \left(2.2\times10^{-10}\right)\left(4.4\times10^{-10}\right)^{2}\\ &= 4.3\times10^{-29} \end{aligned} \] The Ksp value for Hg2I2 is \(4.3\times10^{-29}\).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hg2I2
Mercury(I) iodide, also known as \(\text{Hg}_2\text{I}_2\), is a fascinating compound. It's an example of a sparingly soluble salt, which means it doesn't dissolve easily in water. Instead, most of it remains as a solid when mixed with water.
When \(\text{Hg}_2\text{I}_2\) is dissolved to some extent, it can separate into its ions, \(\text{Hg}_2^{2+}\) and \(\text{I}^-\). These ions are crucial when it comes to understanding the solubility product constant \(K_{\text{sp}}\).
  • \(\text{Hg}_2\text{I}_2\) is often used in equilibrium studies to explore how substances dissolve.
  • It's known for having distinct properties, such as changing color from yellow to red when heated, indicating a change in its crystalline structure.
Understanding \(\text{Hg}_2\text{I}_2\) is key to mastering the concepts of solubility, equilibrium, and chemical calculations.
Radioactivity
Radioactivity involves the decay of unstable atomic nuclei, which releases energy in the form of particles or electromagnetic waves. In the context of this exercise, the isotope \(^{131}\text{I}\) was used.
\(^{131}\text{I}\) is a radioactive isotope of iodine often used in scientific studies because its decay can be measured as counts per minute. These measurements help determine the concentration of ions in solutions.
  • The use of \(^{131}\text{I}\) allows for precise determination of concentration by tracking radioactive decay.
  • Radioactive isotopes are helpful in tracing reactions and calculating exact amounts of substances present.
This exercise uses the radioactivity of iodine to understand chemical reactions and equilibrium processes better.
Chemical Equilibrium
Chemical equilibrium refers to a state in which the rate of the forward reaction matches the rate of the reverse reaction. For \(\text{Hg}_2\text{I}_2\), equilibrium is reached when the solid hasn't dissolved further, but some ions are present in the solution.
In our case, \(\text{Hg}_2\text{I}_2(s) \rightleftharpoons \text{Hg}_2^{2+}(aq) + 2\text{I}^-(aq)\), this equilibrium is critical for solving the problem.
  • At equilibrium, the concentration of each ion doesn't change, even though the reactions are ongoing.
  • This balance is expressed using the solubility product constant \(K_{\text{sp}}\), which quantifies the concentration of ions in solution.
Understanding chemical equilibrium is crucial for predicting how much of a compound will dissolve in a solution.
Concentration Calculation
Concentration calculations involve determining how much of a substance is present in a given volume of solution. In this exercise, the concentration of ions from \(\text{Hg}_2\text{I}_2\) was calculated.
By using radioactivity data, we first determined the moles of \(^{131}\text{I}\) and then used that to find the concentration of \(\text{I}^-\) ions.
  • Concentration is often expressed in molarity (M), which is moles per liter.
  • Knowing the concentration of \(\text{I}^-\) helped determine the concentration of \(\text{Hg}_2^{2+}\) ions.
Finally, these concentrations were used to calculate the solubility product constant \(K_{\text{sp}}\), allowing us to understand the extent to which \(\text{Hg}_2\text{I}_2\) dissolves in water.

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

Natural uranium is mostly nonfissionable \(^{238} \mathrm{U}\) it contains only about 0.7\(\%\) of fissionable \(^{235} \mathrm{U}\) . For uranium to be useful as a nuclear fuel, the relative amount of \(^{235} \mathrm{U}\) must be increased to about 3\(\% .\) This is accomplished through a gas diffusion process. In the diffusion process, natural uranium reacts with fluorine to form a mixture of \(^{238} \mathrm{UF}_{6}(g)\) and 235 \(\mathrm{UF}_{6}(g) .\) The fluoride mixture is then enriched through a multistage diffusion process to produce a 3\(\%^{235} \mathrm{U}\) nuclear fuel. The diffusion process utilizes Graham's law of effusion (see Chapter 5,Section 5.7). Explain how Graham's law of effusion allows natural uranium to be enriched by the gaseous diffusion process.

Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \(^{24} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a 5.00 -g sample of \(^{241} \mathrm{Am}\) ?

There is a trend in the United States toward using coal-fired power plants to generate electricity rather than building new nuclear fission power plants. Is the use of coal-fired power plants without risk? Make a list of the risks to society from the use of each type of power plant.

Which do you think would be the greater health hazard: the release of a radioactive nuclide of Sr or a radioactive nuclide of Xe into the environment? Assume the amount of radioactivity is the same in each case. Explain your answer on the basis of the chemical properties of Sr and Xe. Why are the chemical properties of a radioactive substance important in assessing its potential health hazards?

A living plant contains approximately the same fraction of carbon-14 4 as in atmospheric carbon dioxide. Assuming that the observed rate of decay of carbon-14 4 from a living plant is 13.6 counts per minute per gram of carbon, how many counts per minute per gram of carbon will be measured from a \(15,000\) -year-old sample? Will radiocarbon dating work well for small samples of 10 \(\mathrm{mg}\) or less? (For \(^{14} \mathrm{C}, t_{1 / 2}=5730\) years.)

See all solutions

Recommended explanations on Chemistry Textbooks

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.