Answer to Question #237095 in General Chemistry for Nana

Question

The K_sp value for magnesium carbonate, MgCO₃, is 5.9 x 10^-6. If 2.50 g of magnesium carbonate is placed in 1.00 x 10² mL of water, how much (in g) magnesium carbonate will dissolve? (Please provide your answer with 2 significant figures.)

Answer 1

$MgCO_3 \leftrightarrows Mg^{2+} + CO_3^{2-}$

number of moles of MgCO₃ = mass/molar mass = $\dfrac{2.50g}{84.3g/mol} =$ 0.02966 moles

volume = 1.00 × 10² mL = 100 mL = 0.100 L

molarity = 0.02966/0.100 = 0.2966 M

$K_{sp} = [Mg^{2+}][CO_3^{2-}]\\ 5.9 × 10^{-6}= s²\\ s = \sqrt{5.9 × 10^{-6}} = 0.00243M$

Since, molar mass of magnesium carbonate = 84.32 g/mol.

Grams of magnesium carbonate dissolved in per 1 liter solution (mass concentration) $=0.00243\ M × 84.32\ g/mol=0.2049 g/L$

mass dissolved = 0.2049g/L × 0.100L = 0.02049g