본문 바로가기
일반화학/[16장] 산-염기 평형

0.200 M HCl 10.00 mL + 0.250 M NaOH. 2.00 mL

by 영원파란 2023. 1. 2.

AdSense@ywpop

0.200 M HCl 10.00 mL + 0.250 M NaOH. 2.00 mL

 

 

In a titration, 10.00 mL of 0.200 M HCl(aq)

is titrated with standardized 0.250 M NaOH(aq).

What is the amount of unreacted HCl(aq) and the pH of the solution

after the following volumes of NaOH(aq) have been added?

b) 2.00 mL

 

---------------------------------------------------

▶ 참고: 강산-강염기 적정

[ https://ywpop.tistory.com/2732 ]

---------------------------------------------------

 

(0.200 mol/L) (10.00/1000 L) = 0.00200 mol HCl

---> 처음 산의 mol수

 

 

 

(0.250 mol/L) (2.00/1000 L) = 0.000500 mol NaOH

---> 가한 염기의 mol수

 

 

 

0.00200 – 0.000500 = 0.00150 mol HCl

---> unreacted HCl

 

 

 

전체 용액의 부피 = 10.00 + 2.00 = 12.00 mL

 

 

 

[HCl] = 0.00150 mol / (12.00/1000 L)

= 0.125 M

 

 

 

[H^+] = [HCl] = 0.125 M 이므로,

( 참고 https://ywpop.tistory.com/12475 )

 

pH = –log[H^+]

= –log(0.125) = 0.903

 

 

반응형
그리드형(광고전용)

댓글