Asked by Linda
A student titrates a 25.00 mL sample of HCl. If 20.50 mL of 0.175 M of KOH is required to reach the end point, what is the concentration of the HCl?
25.0 mL HCl x 1L/1000 mL x 0.175 M/1 L = 0.004375 mol HCl
0.004375 mol HCl/20.50 mL KOH x 1000 mL/1 L = 0.213 M HCl
(I'm thinking maybe I have the 20.50 mL or KOH and the 25.0 mL HCl in the wrong places, they need to be switched?)
25.0 mL HCl x 1L/1000 mL x 0.175 M/1 L = 0.004375 mol HCl
0.004375 mol HCl/20.50 mL KOH x 1000 mL/1 L = 0.213 M HCl
(I'm thinking maybe I have the 20.50 mL or KOH and the 25.0 mL HCl in the wrong places, they need to be switched?)
Answers
Answered by
bobpursley
You have it messed up. Learn to do it this way.
NormalityAcid=NormalBAse
If you don't know how to convert M to N, learn that. In this case, they are the same.
NormalityHCL*volumeAcit=.175*20.50
MolarityHCL=.175*20.5ml*M/25.0ml
and you have it. it is not your answer above.
NormalityAcid=NormalBAse
If you don't know how to convert M to N, learn that. In this case, they are the same.
NormalityHCL*volumeAcit=.175*20.50
MolarityHCL=.175*20.5ml*M/25.0ml
and you have it. it is not your answer above.