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.
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.
To solve this question, we can use the concept of stoichiometry in a titration.
First, we need to understand the relationship between the unknown solution and the standard solution. We know that 10.0 mL of the unknown solution requires close to 20 mL of the standard solution for each trial.
We can set up a proportion by equating the volumes of the unknown and standard solutions:
(10.0 mL of unknown solution) / (close to 20 mL of standard solution) = (20 mL of unknown solution) / (x mL of standard solution)
By cross-multiplying, we get:
10.0 mL * x mL of standard solution = (close to 20 mL) * (20 mL)
Simplifying the equation gives:
10.0x = 400
Now, we can solve for x:
x = 400 / 10.0
x = 40 mL
Therefore, approximately 40 mL of the standard solution would be needed to titrate 20 mL of the solution with the unknown concentration.