10 mL*x = 20*y
20*x = ?y. Wouldn't that be 40 mL of the std?
If that's confusing to you (and the whole problem is confusing), just make up a number for the standard solution. Let's call that 0.1M
So 10mLunk x ?M = 20mLstd x 0.1M
Solve for ?M unk = (20x0.1/10) = 0.2M for the unknown. Now we reverse that.
20 mL unk x 0.2 = ?mLstd x 0.1M
?mL std = 20unk x 0.2/0.1 = 40 mL std.
I really need help with this question.
If 2 trials using 10.0 mL of a solution with an unknown concentration were titrated using close to 20mL of a standard solution each trial, what approximate volume of the standard solution would be needed to titrate 20mL of the solution wit the unknown concentration.
1 answer