20.00 mL Ba(OH)2 + 0.245 M HCl 27.15 mL

20.00 mL Ba(OH)2 + 0.245 M HCl 27.15 mL

 

 

A 20.00 mL sample of a Ba(OH)2 solution is titrated with 0.245 M HCl.

If 27.15 mL of HCl is required, what is the molarity of the Ba(OH)2 solution?

 

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aMV = bM’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(1) (0.245 M) (27.15 mL) = (2) (? M) (20.00 mL)

 

? = [(1) (0.245) (27.15)] / [(2) (20.00)]

= 0.166 M

 

 

 

답: 0.166 M