Question Video: Calculating the Solubility Product of Barium Carbonate at Room Temperature | Nagwa Question Video: Calculating the Solubility Product of Barium Carbonate at Room Temperature | Nagwa

# Question Video: Calculating the Solubility Product of Barium Carbonate at Room Temperature Chemistry • Third Year of Secondary School

The solubility of barium carbonate (BaCO₃) at room temperature is 5.08 × 10⁻⁵ mol⋅dm⁻³. What is the solubility product of barium carbonate at this temperature? Give your answer in scientific notation to 2 decimal places.

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### Video Transcript

The solubility of barium carbonate, BaCO3, at room temperature is 5.08 times 10 to the negative five moles per cubic decimeter. What is the solubility product of barium carbonate at this temperature? Give your answer in scientific notation to two decimal places. (A) 2.58 times 10 to the negative nine moles squared times decimeters to the negative six. (B) 1.31 times 10 to the negative 13 moles squared times decimeters to the negative six. (C) 1.02 times 10 to the negative four moles squared times decimeters to the negative six. (D) 5.16 times 10 to the negative nine moles squared times decimeters to the negative six. Or (E) 2.54 times 10 to the negative five moles squared times decimeters to the negative six.

In this question, we want to find the solubility product, abbreviated K sp, of barium carbonate. The solubility product of a compound is the product of the concentrations of the ions in a saturated solution raised to the power of their respective stoichiometric coefficients. We can look at this using an equilibrium equation of the generic solid compound, MA, dissociating into its constituent ions.

The lowercase letters represent the stoichiometric coefficients of the ions in solution. So, the K sp would be written as shown. We use brackets to denote concentration, which we can measure in units of moles per cubic decimeter. We are given from the problem the solubility of barium carbonate at room temperature measured in moles per cubic decimeter. We can use the generic equation to write a balanced equilibrium equation for barium carbonate.

Barium carbonate would be in equilibrium with barium cations and carbonate anions. Using this equation, we can write the K sp expression. At equilibrium, all species are stoichiometrically equivalent. As each unit of barium carbonate dissolves, one barium ion and one carbonate ion are produced. Per cubic decimeter of water, 5.08 times 10 to the negative five moles of barium carbonate can dissolve. Thus, the concentrations of the produced ions will also be 5.08 times 10 to the negative five moles per cubic decimeter.

We can substitute these concentrations into the K sp equation. We can multiply these concentrations together to find the numerical value of the solubility product. To find the units, we multiply moles times moles to get moles squared, and decimeters to the negative three times decimeters to the negative three to get decimeters to the negative six. Our answer is already in scientific notation. But we need to round it to two decimal places, to get 2.58 times 10 to the negative nine moles squared times decimeters to the negative six.

Therefore, the solubility product of barium carbonate at room temperature is answer choice (A), 2.58 times 10 to the negative nine moles squared times decimeters to the negative six.

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