ahs4n MS0 10+ Year Member Joined Mar 7, 2009 Messages 74 Reaction score 2 Points 4,551 Pre-Medical May 6, 2009 #1 Advertisement - Members don't see this ad What would be the pH if .38 g of Fe+3 reacted with 25 L of water assuming only the following equation contributed to pH. Fe+3 + H2O yields 3H+ + Fe(OH)3
Advertisement - Members don't see this ad What would be the pH if .38 g of Fe+3 reacted with 25 L of water assuming only the following equation contributed to pH. Fe+3 + H2O yields 3H+ + Fe(OH)3
ahs4n MS0 10+ Year Member Joined Mar 7, 2009 Messages 74 Reaction score 2 Points 4,551 Pre-Medical May 6, 2009 #2 Sorry, but I was wondering why the beginning step necessitated that you divide the .38 grams by the 25 L of water Upvote 0 Downvote
Sorry, but I was wondering why the beginning step necessitated that you divide the .38 grams by the 25 L of water
L loveoforganic -Account Deactivated- 10+ Year Member Joined Jan 30, 2009 Messages 4,218 Reaction score 14 Points 4,621 May 6, 2009 #3 pH = -log[(0.38 grams Fe3+)*(1 mole Fe3+/55.845 grams Fe3+)*(3 mole H+/1 mole Fe3+) / (25 L)] You have to divide by 25 L because molarity (concentration) is given in units of moles / 1 L, not moles per 25 L Upvote 0 Downvote
pH = -log[(0.38 grams Fe3+)*(1 mole Fe3+/55.845 grams Fe3+)*(3 mole H+/1 mole Fe3+) / (25 L)] You have to divide by 25 L because molarity (concentration) is given in units of moles / 1 L, not moles per 25 L
funsmith Full Member 10+ Year Member Joined May 6, 2009 Messages 20 Reaction score 0 Points 0 Pre-Medical May 6, 2009 #4 Very nice! Upvote 0 Downvote