0.5440 M H3PO4 55.25 mL + 0.2450 M KOH

0.5440 M H3PO4 55.25 mL + 0.2450 M KOH

 

 

0.5440 M H3PO4 용액 55.25 mL를 완전히 중화하려면

0.2450 M KOH 용액 몇 mL가 필요한가?

 

 

What volume (mL) of a 0.2450 M KOH(aq) solution is required

to completely neutralize 55.25 mL of a 0.5440 M H3PO4(aq) solution?

 

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(0.5440 mol/L) (55.25/1000 L) = 0.030056 mol H3PO4

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

H3PO4 + 3KOH → K3PO4 + 3H2O

 

H3PO4 : KOH = 1 : 3 계수비(= 몰수비) 이므로,

H3PO4 : KOH = 1 : 3 = 0.030056 mol : ? mol

? = 3 × 0.030056 = 0.090168 mol KOH

 

 

 

0.090168 mol / (0.2450 mol/L) = 0.3680327 L

= 368.0 mL KOH

 

 

 

 

또는

> H3PO4의 당량수 = 3 eq/mol = a

> KOH의 당량수 = 1 eq/mol = b

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

 

 

 

aMV = bM’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(3) (0.5440 M) (55.25 mL) = (1) (0.2450 M) (? mL)

? = [(3) (0.5440) (55.25)] / [(1) (0.2450)] = 368.0 mL

 

 

 

답: 368.0 mL