FINDING UNKNOWN CONCENTRATIONS OF SOLUTIONS THROUGH TITRATIONS
Titration is also used to find the concentration of an unknown solution. But, how do you do this?! The first step is to use this equation:
Molarity of acid * Volume of acid = Molarity base * V of base ---> Macid * Vacid = Mbase * Vbase
Then next step is to fill in what you know. The molarity of either an acid or base will be provided, and to find the volume of the acid or base being dripped, you subtract final burette readings from the initial. Then, plug in the volume of the acid or base in the flask. What's left from there is just simple algebra!
Still don't quite understand? Well, here's an example using the sample experiment information.
Find the molarity of 10 cc of NaOH that has been neutralized by 1M of 10 cc of HCl.
Molarity of acid = 1
Molarity of base = X (unknown)
Volume of acid = 10 cc
Volume of base = 10 cc
Macid * Vacid = Mbase * Vbase ---> 1M * 10 cc = X * 10 cc =
1M * 10 cc/10cc = X
1 M = X
Answer: The molarity of 10 cc of NaOH is 1M.
Molarity of acid * Volume of acid = Molarity base * V of base ---> Macid * Vacid = Mbase * Vbase
Then next step is to fill in what you know. The molarity of either an acid or base will be provided, and to find the volume of the acid or base being dripped, you subtract final burette readings from the initial. Then, plug in the volume of the acid or base in the flask. What's left from there is just simple algebra!
Still don't quite understand? Well, here's an example using the sample experiment information.
Find the molarity of 10 cc of NaOH that has been neutralized by 1M of 10 cc of HCl.
Molarity of acid = 1
Molarity of base = X (unknown)
Volume of acid = 10 cc
Volume of base = 10 cc
Macid * Vacid = Mbase * Vbase ---> 1M * 10 cc = X * 10 cc =
1M * 10 cc/10cc = X
1 M = X
Answer: The molarity of 10 cc of NaOH is 1M.
******NOTE: In this example, the volume of the acid was found out by subtracting final burette readings from the initial burette readings.*********