QuestionApril 21, 2025

At a certain temperature the solubility of strontium arsenate, Sr_(3)(AsO_(4))_(2) is 0.0530g/L What is the K_(sp) of this salt at this temperature? K_(sp)=square

At a certain temperature the solubility of strontium arsenate, Sr_(3)(AsO_(4))_(2) is 0.0530g/L What is the K_(sp) of this salt at this temperature? K_(sp)=square
At a certain temperature the solubility of strontium arsenate, Sr_(3)(AsO_(4))_(2) is 0.0530g/L
What is the K_(sp) of this salt at this temperature?
K_(sp)=square

Solution
4.0(197 votes)

Answer

K_{sp} = 1.09 \times 10^{-18} Explanation 1. Calculate molar mass of Sr_{3}(AsO_{4})_{2} Molar mass = 3 \times 87.62 + 2 \times (74.92 + 4 \times 16.00) = 3 \times 87.62 + 2 \times 238.84 = 452.8 \, g/mol 2. Convert solubility to molarity Solubility in moles/L = \frac{0.0530 \, g/L}{452.8 \, g/mol} = 1.17 \times 10^{-4} \, mol/L 3. Write dissociation equation Sr_{3}(AsO_{4})_{2} \rightarrow 3Sr^{2+} + 2AsO_{4}^{3-} 4. Determine ion concentrations [Sr^{2+}] = 3 \times 1.17 \times 10^{-4} = 3.51 \times 10^{-4} \, mol/L [AsO_{4}^{3-}] = 2 \times 1.17 \times 10^{-4} = 2.34 \times 10^{-4} \, mol/L 5. Calculate K_{sp} **K_{sp} = [Sr^{2+}]^3 \cdot [AsO_{4}^{3-}]^2** K_{sp} = (3.51 \times 10^{-4})^3 \cdot (2.34 \times 10^{-4})^2 = 1.09 \times 10^{-18}

Explanation

1. Calculate molar mass of $Sr_{3}(AsO_{4})_{2}$<br /> Molar mass = $3 \times 87.62 + 2 \times (74.92 + 4 \times 16.00) = 3 \times 87.62 + 2 \times 238.84 = 452.8 \, g/mol$<br /><br />2. Convert solubility to molarity<br /> Solubility in moles/L = $\frac{0.0530 \, g/L}{452.8 \, g/mol} = 1.17 \times 10^{-4} \, mol/L$<br /><br />3. Write dissociation equation<br /> $Sr_{3}(AsO_{4})_{2} \rightarrow 3Sr^{2+} + 2AsO_{4}^{3-}$<br /><br />4. Determine ion concentrations<br /> $[Sr^{2+}] = 3 \times 1.17 \times 10^{-4} = 3.51 \times 10^{-4} \, mol/L$<br /> $[AsO_{4}^{3-}] = 2 \times 1.17 \times 10^{-4} = 2.34 \times 10^{-4} \, mol/L$<br /><br />5. Calculate $K_{sp}$<br /> **$K_{sp} = [Sr^{2+}]^3 \cdot [AsO_{4}^{3-}]^2$**<br /> $K_{sp} = (3.51 \times 10^{-4})^3 \cdot (2.34 \times 10^{-4})^2 = 1.09 \times 10^{-18}$
Click to rate:

Similar Questions