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Chapter 13: Problem 1

If the \(\mathrm{K}_{\text {sp }}\) of \(\mathrm{CaSO}_4\) is calculated to be \(4.93 \times 10^{-5}\) at \(25^{\circ} \mathrm{C}\), what is the minimum amount of \(\mathrm{CaSO}_4\) that can be added to \(3.75 \times 10^5 \mathrm{~L}\) of water to create a saturated solution? A. \(2.63 \times 10^3\) grams B. \(3.58 \times 10^5\) grams C. \(7.16 \times 10^5\) grams D. \(2.52 \times 10^3\) grams

Short Answer

Expert verified
The minimum amount of \( \text{CaSO}_4 \) that can be added is approximately \3.58 \times 10^5 \text{grams}\) (Option B).

Step by step solution

01

Understand the dissociation reaction

First, write the dissociation reaction for \(\text{CaSO}_4\) in water: \[ \text{CaSO}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]
02

Establish the solubility product expression (Ksp)

The solubility product constant expression is given by: \[ \text{K}_{\text {sp}} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \] Since both ions are produced in a 1:1 ratio, we can let \( s \) be the solubility of \( \text{CaSO}_4 \) in mol/L, then \[ [\text{Ca}^{2+}] = s \] and \[ [\text{SO}_4^{2-}] = s \].
03

Relate solubility to Ksp

Substitute the solubility \(s \) into the \(\text{K}_{\text {sp}}\) expression: \[ \text{K}_{\text {sp}} = s \times s = s^2 \] Since \( \text{K}_{\text {sp}} = 4.93 \times 10^{-5} \), we get \[ s^2 = 4.93 \times 10^{-5} \]
04

Solve for solubility (s)

To find the solubility \( s \,\) take the square root of both sides of the equation: \[ s = \sqrt{4.93 \times 10^{-5}} \approx 7.02 \times 10^{-3} \text{mol/L} \]
05

Convert solubility to grams

To find the grams of \( \text{CaSO}_4 \) that can be dissolved in 1 liter of water, use the molar mass of \( \text{CaSO}_4 \): \[ \text{Molar mass of CaSO}_4 = 40.08 + 32.07 + 4(16.00) = 136.14 \text{g/mol} \] Thus, \[ 7.02 \times 10^{-3} \text{mol/L} \times 136.14 \text{g/mol} \approx 0.955 \text{g/L} \]
06

Find the total amount for the given volume

Finally, multiply the solubility in grams per liter by the total volume of water: \[ 0.955 \text{g/L} \times 3.75 \times 10^5 \text{L} \approx 3.58 \times 10^5 \text{grams} \]

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Key Concepts

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

solubility product constant
Understanding the solubility product constant, or \text{K}_{\text {sp}}\, is crucial in solving problems involving slightly soluble salts like \text{CaSO}_4\. The \text{K}_{\text {sp}}\ quantifies the solubility of a compound under equilibrium conditions. For \text{CaSO}_4\, the \text{K}_{\text {sp}}\ value is \4.93 \times 10^{-5}\, meaning this is the product of the molar concentrations of \(\text{Ca}^{2+}\) and \(\text{SO}_4^{2-}\) ions in a saturated solution. Keep in mind:
  • Higher \text{K}_{\text {sp}}\texpresses a more soluble compound.
  • Lower \text{K}_{\text {sp}}\texpresses a less soluble compound.

This specific equilibrium deals with sparingly soluble salts, not easily dissolving in water.
dissociation reaction
The dissociation reaction is fundamental in determining how a salt dissolves in water. For \text{CaSO}_4,\ the reaction can be written as:
\[ \text{CaSO}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]
This demonstrates that each mole of \text{CaSO}_4\ solid produces one mole of \text{Ca}^{2+}\ ions and one mole of \text{SO}_4^{2-}\ ions. Notice the 1:1 stoichiometric ratio:
  • When the solid dissolves, it splits into two different ions.
  • Each ion increases in concentration by an equal amount.

Recognizing this dissociation helps translate the problem from grams to moles, and set up the \text{K}_{\text {sp}}\ equation correctly.
molar mass calculation
To convert between grams and moles, you need the molar mass of the compound. For \text{CaSO}_4,\ the calculation is:
\[ \text{Molar mass of CaSO}_4 = 40.08 \text{(Ca)} + 32.07 \text{(S)} + 4(16.00 \text{(O)}) = 136.14 \text{g/mol} \]
This process involves finding the atomic masses of calcium (Ca), sulfur (S), and oxygen (O):
  • 40.08 g/mol for Ca.
  • 32.07 g/mol for S.
  • 16.00 g/mol for each of the four oxygen atoms.

Summing these gives 136.14 g/mol. This value allows you to convert moles to grams, and thus find the total mass of \text{CaSO}_4\ that can dissolve in the water.
solubility in water
Solubility in water tells you how much of the solid can dissolve to form a saturated solution. For \text{CaSO}_4,\ the solubility can be derived from the \text{K}_{\text {sp}}\ and the dissociation reaction. From \text{K}_{\text {sp}} = 4.93 \times 10^{-5},\ we solve:
\[ s = \text{solubility} = \text{ \sqrt{4.93 \times 10^{-5}}} \approx 7.02 \times 10^{-3} \text{mol/L} \]
  • This tells us the maximum concentration of \text{CaSO}_4\ ions that water can dissolve per liter.
  • To convert this solubility to grams, multiply by the molar mass (136.14 g/mol).
Thus, 7.02 \times 10^{-3} mol/L * 136.14 g/mol = 0.955 g/L. This means 0.955 grams of \<\text{CaSO}_4\ dissolves in each liter of water. By multiplying this by the given volume (3.75 \times 10^5 L), the amount of \text{CaSO}_4\ added to water to reach saturation is approximately 3.58 \times 10^5 grams.

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