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Chapter 17: Problem 108

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 (Ksp) of strontium sulfate at \(25^{\circ} \mathrm{C}\) is \(1.25 \times 10^{-6}\).

Step by step solution

01

Write down the osmotic pressure equation

The osmotic pressure equation is given by: \(螤 = cRT\) where 螤 is the osmotic pressure, c is the molar concentration of the solute, R is the ideal gas constant, and T is the absolute temperature in Kelvin.
02

Write down the Van't Hoff equation relating osmotic pressure and molar concentration

The Van't Hoff equation is given by: \(螤 = c_i 脳 R 脳 T\) where c_i is the effective concentration of the solute particles.
03

Write down the balanced chemical equation for strontium sulfate dissolution

The balanced chemical equation for the dissolution of strontium sulfate is: \(SrSO_4(s) 鈫� Sr^{2+}(aq) + SO^{2-}_4(aq)\)
04

Write down the solubility product (Ksp) equation for strontium sulfate

The solubility product (Ksp) equation for strontium sulfate is: \(K_{sp} = [Sr^{2+}] 脳 [SO^{2-}_4]\)
05

Convert the given temperature to Kelvin and determine the ideal gas constant value

To convert the given temperature, 25掳C, to Kelvin, add 273.15: \(25掳C + 273.15 = 298.15K\) The value of the ideal gas constant, R, is 0.0821 L atm/mol K.
06

Calculate the molar concentration of the solute using the osmotic pressure equation

Given the osmotic pressure (21 torr), convert the pressure to atm: \(1 Torr = \frac{1}{760} atm\) \(21 Torr = \frac{21}{760} atm = 0.02763 atm\) Now, we can calculate the molar concentration (c) using the osmotic pressure equation: 螤 = cRT where 螤 = 0.02763 atm, R = 0.0821 L atm/mol K, and T = 298.15 K. Solve for c: \(c = \frac{螤}{RT} = \frac{0.02763}{(0.0821)(298.15)} = 1.12 脳 10^{-3} M\)
07

Calculate the solubility product (Ksp) using the molar concentration

Since the dissolution results in a 1:1 ratio of Sr虏鈦� and SO鈧劼测伝 ions, the molar concentration for each ion is also 1.12 脳 10^{-3} M. Using the solubility product equation: \(K_{sp} = [Sr^{2+}] 脳 [SO^{2-}_4] = (1.12 脳 10^{-3})(1.12 脳 10^{-3}) = 1.25 脳 10^{-6}\) The solubility product of strontium sulfate at 25掳C is 1.25 脳 10^{-6}.

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Most popular questions from this chapter

A hypothetical weak acid, HA, was combined with \(\mathrm{NaOH}\) in the following proportions: \(0.20 \mathrm{~mol}\) of \(\mathrm{HA}\), \(0.080 \mathrm{~mol}\) of \(\mathrm{NaOH}\). The mixture was diluted to a total volume of \(1.0 \mathrm{~L}\), and the pH measured. (a) If \(\mathrm{pH}=4.80\), what is the \(\mathrm{p} K_{a}\) of the acid? (b) How many additional moles of \(\mathrm{NaOH}\) should be added to the solution to increase the \(\mathrm{pH}\) to \(5.00 ?\)

(a) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood of pH \(7.4\) ? (b) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in an exhausted marathon runner whose blood pH is 7.1?

Predict whether the equivalence point of each of the following titrations is below, above, or at \(\mathrm{pH} 7\) : (a) \(\mathrm{NaHCO}_{3}\) titrated with \(\mathrm{NaOH}\), (b) \(\mathrm{NH}_{3}\) titrated with \(\mathrm{HCl}\), (c) KOH titrated with HBr.

A \(30.0-\mathrm{mL}\) sample of \(0.150 \mathrm{M} \mathrm{KOH}\) is titrated with \(0.125 \mathrm{M} \mathrm{HClO}_{4}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of acid have been added: (a) \(30.0 \mathrm{~mL}\), (b) \(35.0 \mathrm{~mL}\), (c) \(36.0 \mathrm{~mL}\), (d) \(37.0 \mathrm{~mL}\), (e) \(40.0 \mathrm{~mL}\).

What is the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of water saturated with \(\mathrm{CO}_{2}\) at a partial pressure of \(1.10 \mathrm{~atm}\) ? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-2} \mathrm{~mol} / \mathrm{L}-\mathrm{atm}\). The \(\mathrm{CO}_{2}\) is an acidic oxide, reacting with \(\mathrm{H}_{2} \mathrm{O}\) to form \(\mathrm{H}_{2} \mathrm{CO}_{3}\).
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