Answer to Question #237095 in General Chemistry for Nana

Question #237095

The Ksp 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.)




1
Expert's answer
2021-09-15T02:41:27-0400

MgCO3Mg2++CO32MgCO_3 \leftrightarrows Mg^{2+} + CO_3^{2-}


number of moles of MgCO3 = mass/molar mass = 2.50g84.3g/mol=\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


Ksp=[Mg2+][CO32]5.9×106=s²s=5.9×106=0.00243MK_{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=0.2049g/L=0.00243\ M × 84.32\ g/mol=0.2049 g/L


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


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