A 250 mL solution contains 0.50 M of Calcium ions (Ca2*) and 0.75 M of Magnesium ions (Mg2 ). What amount in moles of sodium carbonate (NazCO3) must be added to completely precipitate the Calcium ions?CaCOs Ksp = 3.8x10\deg ; MgCO3 Ksp= 6.8x10-6
Added by Tom-S E.
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To determine the amount in moles of sodium carbonate (Na2CO3) needed to completely precipitate the calcium ions (CaĀ²āŗ) from the solution, we can follow these steps: ** Show moreā¦
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David C.
If 18.0 mL of 2.24Ć10^-4 M Mg(NO3)2 are added to 22.0 mL of 5.00Ć10^-5 M Na2CO3, will solid MgCO3 (Ksp = 4.0Ć10^-5) precipitate? If a precipitate will not form, what carbonate ion concentration will cause a precipitate of magnesium carbonate to form? If a precipitate will form, what is the minimum [CO3^2-] that could have been present without initiating precipitation? Assume the total volume used in the above example.
B2. A chemist mixes equal volumes of an aqueous solution of CaCl2 (1.34 Ć 10ā4 M) and an aqueous solution of Na2CO3 (1.34 Ć 10ā4 M) to prepare a saturated solution of calcium carbonate (no precipitate is observed). (a) Calculate the concentration of each of the ions in this saturated solution (assuming 100% dissociation). [Ca2+] = [Na+] = [Clā] = [CO32ā] = (b) Assuming that the chemist has calculated the molar solubility for calcium carbonate accurately, write the expression for the solubility product and calculate the value for Ksp for calcium carbonate.
Susan H.
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