The solubility of \(\mathrm{Sr}(\mathrm{OH})_{2}\) at \(298 \mathrm{~K}\) is \(19.23 \mathrm{~g} / \mathrm{L}\) of solution. Calculate the concentrations of strontium and hydroxyl ions and the \(\mathrm{pH}\) of the solution.

Short Answer

Expert verified
The concentration of \(\mathrm{Sr}^{2+}\) is 0.158 M, the concentration of \(\mathrm{OH}^{-}\) ions is 0.316 M, and the \(\mathrm{pH}\) of the solution can be found using \(\mathrm{pH} = 14 - \mathrm{pOH}\), where \(\mathrm{pOH}\) is calculated using \(\mathrm{pOH} = -\log(0.316)\).

Step by step solution

01

Write down the chemical dissociation

Start by writing down the chemical equation for the dissociation of \(\mathrm{Sr}(\mathrm{OH})_{2}\) in water: \[\mathrm{Sr}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Sr}^{2+}(aq) + 2\mathrm{OH}^{-}(aq)\]. This shows that one mole of strontium hydroxide produces one mole of strontium ion and two moles of hydroxide ions.
02

Calculate molarity of strontium ion

Assuming complete dissociation, the solubility of strontium hydroxide gives the molarity of strontium ion directly. Since the molar mass of \(\mathrm{Sr}(\mathrm{OH})_{2}\) is approximately 121.63 g/mol, the molarity (\(C_{Sr^{2+}}\)) can be calculated using the solubility: \[C_{Sr^{2+}} = \frac{19.23 \mathrm{~g/l}}{121.63 \mathrm{~g/mol}} = 0.158 \mathrm{~mol/l}\].
03

Determine hydroxide ion concentration

From the balanced chemical equation, it is evident that for every mole of \(\mathrm{Sr}(\mathrm{OH})_{2}\), there are two moles of \(\mathrm{OH}^{-}\) ions produced. Therefore, the concentration of hydroxide ions (\(C_{OH^{-}}\)) is twice the concentration of strontium ions: \[C_{OH^{-}} = 2 \times C_{Sr^{2+}} = 2 \times 0.158 \mathrm{~mol/l} = 0.316 \mathrm{~mol/l}\].
04

Calculate the \(\mathrm{pOH}\) of the solution

The \(\mathrm{pOH}\) of the solution can be calculated using the concentration of hydroxide ions: \[\mathrm{pOH} = -\log(C_{OH^{-}}) = -\log(0.316)\].
05

Determine the \(\mathrm{pH}\) of the solution

To find the \(\mathrm{pH}\), use the relationship between \(\mathrm{pH}\) and \(\mathrm{pOH}\) which is \(\mathrm{pH} + \mathrm{pOH} = 14\) at 298 K. Solving for \(\mathrm{pH}\), we get: \[\mathrm{pH} = 14 - \mathrm{pOH} \]. After calculating the \(\mathrm{pOH}\), just subtract it from 14 to find the \(\mathrm{pH}\) of the solution.

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Key Concepts

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

Chemical Dissociation
Understanding chemical dissociation is crucial when dealing with soluble ionic compounds. Dissociation refers to the process where a solid ionic compound separates into its ions when dissolved in water. In the case of Sr(OH)2, the dissociation in water results in strontium ions (Sr2+) and hydroxide ions (OH). The balanced chemical equation is indispensable for predicting the mole ratio in which the ions will be present in the solution.

When improving exercises or solutions, it is helpful to explain that each dissociated compound has a specific ratio of ions that it produces. For Sr(OH)2, one mole yields one mole of Sr2+ and two moles of OH. Ensuring learners grasp this aspect of stoichiometry can eliminate confusion about how many ions result from the dissolution of a given amount of ionic compound.
Molarity
The concept of molarity is a measure of concentration referring to the number of moles of a solute present in one liter of solution. It's a fundamental concept for calculations in chemistry, symbolized by 'M' and formally defined as moles of solute per liters of solution. It is directly used to relate the solubility of a substance to its concentration in a solution. For instance, in this exercise, the molarity of Sr2+ is calculated from the solubility of Sr(OH)2.

Highlighting the explicit relationship between mass, molar mass, and volume when discussing molarity can significantly clarify calculations. Providing examples, such as the molarity of a sugar solution or the concentration of a salt in seawater, can also cement understanding of the topic. It is essential to convey that molarity is not merely an abstract number but directly correlates to how much of a substance exists in a certain volume of liquid, which is a crucial factor in chemical reactions and solubility.
Hydroxide Ion Concentration
The hydroxide ion concentration is paramount when determining the basicity of aqueous solutions. In the context of solubility calculations, once the molarity of the ionic compound is known, the concentration of the hydroxide ions can be derived. As seen in this problem, the dissociation equation outlines that each unit of Sr(OH)2 releases two hydroxide ions, hence the concentration of OH - is double that of the strontium ion concentration.

Improving content around this topic involves emphasizing the connection between the stoichiometry of the dissociation reaction and the resulting ion concentrations. It may be helpful to illustrate this concept by comparing it with more familiar everyday quantities, such as the way doubling a recipe increases the number of servings. Connecting abstract chemical concepts to real-world quantities aids in making these concepts more digestible for students.
pH Calculation
The calculation of pH is a method for expressing the acidity or basicity of a solution on a logarithmic scale. The pH scale typically ranges from 0 to 14, with lower values representing acidic conditions and higher values corresponding to basic conditions. To calculate pH from hydroxide ion concentration, one must first determine the pOH by taking the negative logarithm of the hydroxide ion concentration. Then, since pH and pOH are inversely related and their sum is 14 (at room temperature), one can find the pH.

For clearer content delivery, we can analogize the mathematical relationship between pH and pOH to a see-saw, where one side going up forces the other side down. Moreover, pointing out common pH values, such as those of lemon juice or household bleach, may make this abstract concept more tangible. Lastly, in teaching pH calculations, it's vital to stress the importance of the logarithmic scale, which makes each single unit change in pH correspond to a tenfold change in ion concentration—a counterintuitive but fundamental aspect of understanding pH.

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Most popular questions from this chapter

(a) Suggest a solvent in which aniline acts as strong base. (b) Write equation for the auto ionisation of (i) \(\mathrm{HCOOH}\), (ii) \(\mathrm{NH}_{3}\). (c) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is acid or base and write its conjugate partner and reaction. (d) Write the order of acidic nature of \(\mathrm{HCl}, \mathrm{HCOOH}\) and \(\mathrm{CH}_{3} \mathrm{COOH}\) in (i) \(\mathrm{H}_{2} \mathrm{O}\), (ii) liq. \(\mathrm{NH}_{3}\).

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