0.116 M NaOH 50.0 mL + 0.0750 M HCl 적정. c) 0.10 L

0.116 M NaOH 50.0 mL + 0.0750 M HCl 적정. c) 0.10 L

 

 

Consider the titration of 50.0 mL of 0.116 M NaOH with 0.0750 M HCl.

Calculate the pH after the addition of each of the following volumes of acid:

a) 5.0 mL b) 50 mL c) 0.10 L

 

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▶ 참고: 강산-강염기 적정 [ https://ywpop.tistory.com/2732 ]

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0.116 M NaOH 50.0 mL에 들어있는 NaOH의 몰수

= (0.116 mol/L) (50.0/1000 L) = 0.00580 mol NaOH

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

0.0750 M HCl 0.10 L에 들어있는 HCl의 몰수

= (0.0750 mol/L) (0.10 L) = 0.00750 mol HCl

 

 

 

HCl + NaOH → NaCl + H2O

HCl : NaOH = 1 : 1 반응.

 

 

 

중화 반응 후 남아있는 HCl의 몰수

= 0.00750 – 0.00580 = 0.00170 mol HCl

 

 

 

[HCl] = 0.00170 mol / [(50.0+100)/1000 L]

= 0.0113 M = [H^+]

 

 

 

pH = -log[H^+]

( 참고 https://ywpop.tistory.com/2706 )

 

= -log(0.0113)

= 1.95

 

 

 

 

[키워드] 강산-강염기 적정 기준문서, HCl + NaOH 적정 기준문서