moles HBr = M x L = 0.2M x 0.025mL = 0.005.
moles KOH = M x L = 0.1M x 0.0513 mL = 0.00503
...............HBr + KOH ==> KBr + H2O
initial.....0.005...0.00513..0......0
change.....-0.005..-0.00500.+0.005.+.005
equil.........0.....0.00013..0.005..0.005
It should be clear from the ICE chart that all of the HBr has been used (this is just another limiting reagent problem when we do it this way) and we have an excess of 0.00013 moles KOH left after the reaction. To find M, that is moles/L. You have moles. L = 51.3 mL + 25 mL = ?? L(not mL).
Calculate the OH− concentration after 51.3
mL of the 0.100 M KOH has been added to
25.0 mL of 0.200 M HBr. Assume additive
volumes.
Answer in units of M.
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