150 L concentrated HCl (12.0 M) lake 5.0×10^5 m3 water pH

150 L concentrated HCl (12.0 M) lake 5.0×10^5 m3 water pH

 

 

Due to a highway accident, 150 L of concentrated hydrochloric acid (12.0 M) is released into a lake containing 5.0×10^5 m3 of water. If the pH of this lake was 7.0 prior to the accident, what is the pH of the lake following the accident?

 

 

 

(5.0×10^5 m3) (1000 L / 1 m3)

= 5.0×10^8 L

---> 사고 전, 호수 물의 부피

---> pH 7.0 이므로, “호수 물 = 증류수” 라고 간주.

 

 

 

묽힘법칙. M1V1 = M2V2

( 참고 https://ywpop.tistory.com/2859 )

 

(12.0 M) (150 L) = (? M) (150 + 5.0×10^8 L)

 

? = (12.0) (150) / (150 + 5.0×10^8)

= 3.6×10^(-6) M

---> 사고 후, 호수 속 HCl의 몰농도

 

 

 

[HCl] = [H^+] 이므로,

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

 

pH = –log(3.6×10^(-6))

= 5.44

 

 

 

답: pH = 5.44

 

 

 

 

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