In a titration experiment what would be the result on the concentration of the analyte if the drop counter did not count all the drops?
would it be lower since you would just be calculating the concentration at the same point on the curve just with a lower volume of titrant?
4 answers
M = mols/L so the L would be smaller if you counted fewer drops. That means the M will be ......?
Larger. I dont know what i was thinking. i really overlooked that one. thank you
the drops happen, but are not counted
the fewer the drops to the endpoint, the lower the concentration of the unknown
the concentration would appear to be lower than it actually is
the fewer the drops to the endpoint, the lower the concentration of the unknown
the concentration would appear to be lower than it actually is
Yes, but I believe your reasoning is wrong. The drops actually happen in order to get to the equivalence point BUT you don't count them; therefore, the calculation point (say 10 drops instead of 12 drops) is LOWER which makes M = mols/L higher because the volume number you used in the calculation is lower. Note that your statement "the fewer the drops to the end point" is not what is happening. The drops to the end point stays the same BUT you use a smaller number in the calculation. It is the calculation that screws it up; not the titration itself.
1 answer