aMV = bM’V’. 0.210 M solution of NaOH

aMV = bM’V’. 0.210 M solution of NaOH

 

 

Calculate the volume, in milliliters, of a 0.210 M solution of NaOH that will completely neutralize each of the following.

1. 2.20 mL of a 0.830 M solution of H2SO4

2. 3.80 mL of a 1.27 M solution of HNO3

3. 6.00 mL of a 3.29 M solution of HCl

 

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aMV = bM’V’

acid의 mol수 = base의 mol수

( 식 설명 https://ywpop.tistory.com/4689 )

 

 

 

1. (2) (0.830 M) (2.20 mL) = (1) (0.210 M) (? mL)

? = [(2) (0.830) (2.20)] / [(1) (0.210)] = 17.4 mL

 

 

 

2. (1) (1.27 M) (3.80 mL) = (1) (0.210 M) (? mL)

? = [(1) (1.27) (3.80)] / [(1) (0.210)] = 23.0 mL

 

 

 

3. (1) (3.29 M) (6.00 mL) = (1) (0.210 M) (? mL)

? = [(1) (3.29) (6.00)] / [(1) (0.210)] = 94.0 mL

 

 

 

[ 관련 예제 https://ywpop.tistory.com/search/aMV%20=%20bM’V’ ]