Asked by Carrie
What is the hydroxide ion concentration of a solution prepared by dissolving 0.488 g of Mg(OH)2 in 236 mL of water? How do I solve this???
Answers
Answered by
DrBob222
mols Mg(OH)2 = grams/molar mass
M = mols/L soln. ( I assume the 236 mL of water is 236 mL solution.) I obtained M = about 0.04M but you need to do it more accurately. I'm also assuming that all of the 0.488 g dissolves.
M = about 0.04M
There are 2 OH^-/1 molecule Mg(OH)2; therefore, (OH^-) = twice that or about 0.08M
M = mols/L soln. ( I assume the 236 mL of water is 236 mL solution.) I obtained M = about 0.04M but you need to do it more accurately. I'm also assuming that all of the 0.488 g dissolves.
M = about 0.04M
There are 2 OH^-/1 molecule Mg(OH)2; therefore, (OH^-) = twice that or about 0.08M
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