Question 56 of 69intermediate🔧 ApplyNumerical3 marks

The solubility of Sr(OH)2_2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Correct Answer

[Sr2+^{2+}] = 0.1581 M, [OH−^-] = 0.3162 M, pH = 13.5

Exercise: EXERCISES | Q: 6.58 | (Page 46)
For More Understanding

Explanation

To find the concentrations of the ions, we first convert the given solubility from g/L to mol/L (molar solubility) using the molar mass of Sr(OH)2. Then, using the dissociation stoichiometry of Sr(OH)2, we find the concentrations of Sr2+ and OH-. Finally, we calculate the pOH from the hydroxyl ion concentration and use the relation pH + pOH = 14 to find the pH.

Solution Steps

  1. Step 1: Calculate the molar mass of Sr(OH)2_2 Molar mass of Sr(OH)2_2 = 87.6+2×(16+1)=121.687.6 + 2 \times (16 + 1) = 121.6 g/mol

  2. Step 2: Calculate the molar solubility (S) Solubility in g/L = 19.23 g/L Molar solubility, S=19.23 g/L121.6 g/mol=0.1581S = \frac{19.23 \text{ g/L}}{121.6 \text{ g/mol}} = 0.1581 mol/L

  3. Step 3: Determine the concentrations of strontium and hydroxyl ions The dissociation of Sr(OH)2_2 is: Sr(OH)2→_2 \rightarrow Sr2+^{2+} + 2OH−^- [Sr2+]=S=0.1581[\text{Sr}^{2+}] = S = 0.1581 M [OH−]=2S=2×0.1581=0.3162[\text{OH}^-] = 2S = 2 \times 0.1581 = 0.3162 M

  4. Step 4: Calculate the pOH and pH of the solution

    pOH=−log⁡[OH−]=−log⁡(0.3162)=0.5\text{pOH} = -\log[\text{OH}^-] = -\log(0.3162) = 0.5 pH=14−pOH=14−0.5=13.5\text{pH} = 14 - \text{pOH} = 14 - 0.5 = 13.5
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