0.00491 M Ba(OH)2 24.9 mL 중화에 필요한 0.101 M HNO3의 부피

0.00491 M Ba(OH)2 24.9 mL 중화에 필요한 0.101 M HNO3의 부피

0.00491 M Ba(OH)2 24.9 mL 0.101 M HNO3 vol

 

 

What volume of 0.101 M HNO3 is required

to neutralize each of the following solutions?

b. 24.9 mL of 0.00491 M Ba(OH)2

 

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> HNO3의 당량수 = 1 eq/mol

---> a = 1

 

> Ba(OH)2의 당량수 = 2 eq/mol

---> b = 2

( 참고: 당량수 https://ywpop.tistory.com/4105 )

 

 

 

aMV = bM’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(1) (0.101 M) (? mL) = (2) (0.00491 M) (24.9 mL)

? = [(2) (0.00491) (24.9)] / [(1) (0.101)] = 2.42 mL

 

 

 

답: 2.42 mL