Calculate the solubility of BaCO3 (a) in pure water and (b) in a solution in which [CO32-] = 0.289 M. Solubility in pure water = M Solubility in 0.289 M CO32- = M

the solubility of BaCO₃ in pure  water  and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.

Explanation:

To calculate the solubility of BaCO₃ in:

(a) pure water and (b) in a solution in which [CO₃²⁻] = 0.289 M

The (i.e solubility constant ) for BaCO₃= 2.0 × 10⁻⁹

BaCO₃   → Ba²⁺  + CO₃²⁻

ksp = s × s

s² = ksp

s =

s =

s = 4.472 × 10⁻⁵ M

(b) The solubility of BaCO₃ in a solution in which [CO₃²⁻] = 0.289 M

BaCO₃   → Ba²⁺  + CO₃²⁻

ksp = s × s

2.0 × 10⁻⁹ = s × 0.289

s = 2.0 × 10⁻⁹/0.289

s = 6.9204 × 10⁻⁹ M

Thus, the solubility of BaCO₃ in pure  water  and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.

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explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)

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