The concentration of various cations in seawater in moles per liter are Na+ (0.46M); Mg2+ (0.056M); Ca2+ (0.01M); Al3+ (4x10^-7M); Fe3+ (2x10^-7M)
If enough OH- ion is added to precipitate 50% of the Mg2+, what mass of precipitate will be obtained from one liter or seawater
1L contains 0.056 M = 0.056 moles/L. Grams Mg(OH)2 = 0.056 moles x molar mass = ?? grams. Then take half of that.
thanks so much! just one other quick thing, how would i tell what percentage of the other ions will be precipitated?
~ i know the other answer is virtually all, but how do I come to that conclusion?
To determine the mass of precipitate obtained, we need to first calculate the moles of Mg2+ that will be precipitated. Let's go step by step:
1. Determine the moles of Mg2+ initially present:
Given concentration of Mg2+ in seawater = 0.056 M
Moles of Mg2+ = concentration x volume = 0.056 M x 1 L = 0.056 moles
2. Calculate the moles of Mg2+ that will be precipitated by adding OH- ions:
We know that 50% of the Mg2+ will be precipitated.
Moles of Mg2+ to be precipitated = 50% of 0.056 moles = 0.5 x 0.056 moles = 0.028 moles
3. Determine the balanced chemical equation for the precipitation reaction:
Magnesium ions (Mg2+) react with hydroxide ions (OH-) to form magnesium hydroxide (Mg(OH)2).
The balanced equation is: Mg2+ + 2OH- → Mg(OH)2
4. Using the balanced equation, determine the stoichiometry of the reaction:
1 mole of Mg2+ reacts with 2 moles of OH- ions to form 1 mole of Mg(OH)2.
Therefore, moles of OH- ions required = 2 x moles of Mg2+ to be precipitated = 2 x 0.028 moles = 0.056 moles
5. Calculate the mass of Mg(OH)2 precipitate formed:
To find the mass, we need to know the molar mass of Mg(OH)2.
Molar mass of Mg(OH)2 = (24.3 g/mol for Mg) + 2(16.0 g/mol for O) + 2(1.0 g/mol for H) = 58.3 g/mol
Mass of Mg(OH)2 = moles of Mg(OH)2 x molar mass
= 0.028 moles x 58.3 g/mol
= 1.6324 grams (rounded to four decimal places)
Therefore, when enough OH- ions are added to precipitate 50% of the Mg2+ present in one liter of seawater, approximately 1.6324 grams of magnesium hydroxide precipitate will be obtained.