CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
<em><u>Answer </u></em><em><u>:</u></em><em><u> </u></em><em><u>M</u></em><em><u>etal containers</u></em><em><u> </u></em><em><u>are</u></em><em><u> not</u></em><em><u> </u></em><em><u> </u></em><em><u>used for storing acid</u></em><em><u> because most of the time acid reacts with almost every metal and produces </u></em><em><u>salts</u></em><em><u> or oxid</u></em><em><u>e</u></em><em><u>s</u></em><em><u> </u></em><em><u>which alters the acid characteristics making it useless</u></em><em><u> </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>