Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 constant (Kₛₚ) of strontium sulfate at 25°C is approximately 1.23 × 10⁻⁶.

Step by step solution

01

Write down the expression for osmotic pressure

The osmotic pressure (π) of a solution can be determined using the formula: π = (n/V)RT where n is the amount of solute in moles, V is the volume of the solution in liters, R is the ideal gas constant (0.08206 L atm mol⁻¹ K⁻¹) and T is the temperature in Kelvin.
02

Convert the given osmotic pressure to atm

We are given the osmotic pressure of 21 torr. Since the gas constant R is given in atm, we need to convert the pressure to atm: 1 atm = 760 torr So, 21 torr = (21/760) atm ≈ 0.02763 atm
03

Calculate the molar concentration of the solution

We know π = (n/V)RT. Therefore, n/V (molar concentration) can be calculated as: n/V = π / RT We are given the temperature 25°C, which is 298.15 K in Kelvin. Therefore, n/V can be calculated as: n/V = 0.02763 atm / (0.08206 L atm mol⁻¹ K⁻¹ × 298.15 K) ≈ 0.00111 mol/L
04

Write the solubility product expression for strontium sulfate

Strontium sulfate (SrSO₄) dissociates in water as follows: SrSO₄(s) ⇌ Sr²⁺(aq) + SO₄²⁻(aq) The solubility product constant (Kₛₚ) can be written as: Kₛₚ = [Sr²⁺][SO₄²⁻]
05

Calculate the solubility product constant (Kₛₚ)

Since the saturated solution is at equilibrium, the concentrations of Sr²⁺ and SO₄²⁻ are equal (0.00111 mol/L). Therefore, we can calculate the solubility product constant: Kₛₚ = (0.00111 mol/L)(0.00111 mol/L) ≈ 1.23 × 10⁻⁶ So, the solubility product of strontium sulfate at 25°C is approximately 1.23 × 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 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

Calculate the ratio of \(\left[\mathrm{Ca}^{2+}\right]\) to \(\left[\mathrm{Fe}^{2+}\right]\) in a lake in which the water is in equilibrium with deposits of both \(\mathrm{CaCO}_{3}\) and \(\mathrm{FeCO}_{3}\). Assume that the water is slightly basic and that the hydrolysis of the carbonate ion can therefore be ignored.

Aspirin has the structural formula At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{a}\) for aspirin equals \(3 \times 10^{-5}\). If two aspirin tablets, each having a mass of \(325 \mathrm{mg},\) are dissolved in a full stomach whose volume is \(1 \mathrm{~L}\) and whose \(\mathrm{pH}\) is \(2,\) what percent of the aspirin is in the form of neutral molecules?

Consider a beaker containing a saturated solution of \(\mathrm{Pbl}_{2}\) in equilibrium with undissolved \(\mathrm{Pbl}_{2}(s) .\) (a) If solid KI is added to this solution, will the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Pb}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(\mathrm{I}^{-}\) ions in solution increase or decrease?

Calculate the pH at the equivalence point for titrating \(0.200 \mathrm{M}\) solutions of each of the following bases with \(0.200 \mathrm{M} \mathrm{HBr}\) : (a) sodium hydroxide \((\mathrm{NaOH}),(\mathbf{b})\) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\).

The solubility of \(\mathrm{CaCO}_{3}\) is pH dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{s p}=4.5 \times 10^{-9}\right)\) neglecting the acid-base character of the carbonate ion. (b) Use the \(K_{b}\) expression for the \(\mathrm{CO}_{3}^{2-}\) ion to determine the equilibrium constant for the reaction \(\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\) (c) If we assume that the only sources of \(\mathrm{Ca}^{2+}, \mathrm{HCO}_{3}^{-},\) and \(\mathrm{OH}^{-}\) ions are from the dissolution of \(\mathrm{CaCO}_{3},\) what is the molar solubility of \(\mathrm{CaCO}_{3}\) using the preceding expression? What is the \(\mathrm{pH} ?\) (d) If the \(\mathrm{pH}\) is buffered at 8.2 (as is historically typical for the ocean), what is the molar solubility of \(\mathrm{CaCO}_{3} ?\) (e) If the \(\mathrm{pH}\) is buffered at \(7.5,\) what is the molar solubility of \(\mathrm{CaCO}_{3} ?\) How much does this drop in \(\mathrm{pH}\) increase solubility? solution remains \(0.50 \mathrm{~L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

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.

Sign-up for free