/*! 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 108 The osmotic pressure of a satura... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(25^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The solubility product of strontium sulfate at \(25^{\circ} \mathrm{C}\) is approximately \(1.21 \times 10^{-6}\).

Step by step solution

01

Calculate molar concentration from osmotic pressure

To find the molar concentration, we can use the osmotic pressure equation: \[Π = MRT\] Where: Π = osmotic pressure (21 torr) M = molar concentration of the solution R = gas constant (62.36 L⋅torr/mol⋅K) T = temperature in Kelvin (25°C = 298.15 K) Rewrite the formula for molar concentration M: \[M = \frac{Π}{RT}\] Now plug in the given values and solve for M: \[M = \frac{21 \text{ torr}}{(62.36 \frac{\text{L}\cdot \text{torr}}{\text{mol} \cdot \text{K}})(298.15 \text{ K})} ≈ 0.0011 \text{ mol/L}\]
02

Express equilibrium concentrations in terms of solubility product

Strontium sulfate (SrSO_4) dissociates into ions in water as follows: \[SrSO_4(s) \rightleftharpoons Sr^{2+}(aq) + SO_4^{2-}(aq)\] The solubility product (Ksp) of this reaction is represented by the product of the equilibrium concentrations of the ions: \[K_{sp} = [Sr^{2+}][SO_4^{2-}]\] Since both ions are produced in a 1:1 ratio, their respective equilibrium concentrations will be equal and we can denote it as x. So, we have: \[K_{sp} = x^2\]
03

Calculate the solubility product

Now, we know that the concentration x is approximately 0.0011 mol/L. Substituting this value into the expression for Ksp, we get: \[K_{sp} = (0.0011)^2 ≈ 1.21 \times 10^{-6}\] Therefore, the solubility product of strontium sulfate at 25°C is approximately \(1.21 \times 10^{-6}\).

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.

Understanding Osmotic Pressure
Osmotic pressure is a crucial concept in chemistry that helps us understand how solutions behave when separated by a semipermeable membrane. This pressure arises when solvent molecules move through the membrane from a region of lower concentration to a region of higher concentration. It plays a significant role in maintaining balance across cell membranes and is essential in calculating solution properties, such as concentration.
To calculate osmotic pressure, we use the formula: \( Π = MRT \), where:
  • \( Π \) is the osmotic pressure, measured in units like torr or pascals.
  • \( M \) is the molarity or concentration of the solution.
  • \( R \) is the ideal gas constant.
  • \( T \) is the temperature in Kelvin.
By using this formula, we can calculate the molarity of a solution when other variables are known. In our exercise, the osmotic pressure provided was 21 torr, and using the known values of \( R \) and \( T \), we determined the concentration to be \( 0.0011 \) mol/L.
Equilibrium Concentrations and Solubility Product
Equilibrium concentrations occur when the rate of dissolution and precipitation of a solute in a solution are equal. In the case of sparingly soluble salts, like strontium sulfate, an equilibrium is established between the solid salt and its ions in solution.
For the dissociation of strontium sulfate, the equation is:\[ SrSO_4(s) \rightleftharpoons Sr^{2+}(aq) + SO_4^{2-}(aq) \]This equation shows that each molecule of \( SrSO_4 \) generates one ion of \( Sr^{2+} \) and one ion of \( SO_4^{2-} \).
The solubility product \( K_{sp} \) is determined by the product of the concentrations of these ions at equilibrium.
  • \( K_{sp} = [Sr^{2+}][SO_4^{2-}] \)
Because the ions dissociate in a 1:1 ratio, \([Sr^{2+}]\) and \([SO_4^{2-}]\) are equal, often represented as \( x \). Thus, \( K_{sp} = x^2 \), where \( x \) is the molarity of ions in solution.
Dissociation of Strontium Sulfate
Dissociation refers to the process where a compound splits into its constituent ions or molecules. For strontium sulfate \( (SrSO_4) \), dissociation occurs when it is placed in water. The solid separates into strontium ions \( Sr^{2+} \) and sulfate ions \( SO_4^{2-} \).
This process determines the maximum amount of the salt that can dissolve in water under given conditions, known as its solubility. The low solubility implies a small amount of \( SrSO_4 \) will dissolve before reaching equilibrium. This behavior is used to calculate the solubility product \( K_{sp} \).
  • Each formula unit of \( SrSO_4 \) releases one \( Sr^{2+} \) ion and one \( SO_4^{2-} \) ion upon dissociation.
  • The total concentration of ions in the solution can then be used to determine \( K_{sp} \), which helps assess the salt's solubility under specific conditions.
Understanding the dissociation process and its relationship to equilibrium concentrations is key to determining solubility products.

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

Toothenamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right){ }_{3} \mathrm{OH}\), and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the "Chemistry and Life" box in Section \(17.5\), fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\), whose \(K_{s p}=1.0 \times 10^{-60}\). (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Derive an equation similar to the HendersonHasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.

A buffer solution contains \(0.10 \mathrm{~mol}\) of acetic acid and $0.13 \mathrm{~mol}\( of sodium acetate in \)1.00 \mathrm{~L}$. (a) What is the \(\mathrm{pH}\) of this buffer? (b) What is the \(\mathrm{pH}\) of the buffer after the addition of \(0.02\) mol of KOH? (c) What is the pH of the buffer after the addition of \(0.02 \mathrm{~mol}\) of \(\mathrm{HNO}_{3}\) ?

A sample of \(0.2140 \mathrm{~g}\) of an unkown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M} \mathrm{NaOH}\). The acid required \(27.4 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

A buffer contains a weak acid, \(\mathrm{HX}\), and its conjugate base. The weak acid has a \(\mathrm{pK}_{a}\) of \(4.5\), and the buffer solution has a pH of 4.3. Without doing a calculation, predict whether \([\mathrm{HX}]=\left[\mathrm{X}^{-}\right],[\mathrm{HX}]>\left[\mathrm{X}^{-}\right]\), or \([\mathrm{HX}]<\left[\mathrm{X}^{-}\right]\) Explain. [Section 17.2]

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.