25℃ 500 mL water gram Ba(IO3)2 (487 g/mol)

25℃ 500 mL water gram Ba(IO3)2 (487 g/mol)

 

 

25℃에서 500 mL 물에

몇 그램의 Ba(IO3)2 (487 g/mol)가 녹을 수 있겠는가?

 

 

How many gams of Ba(IO3)2 (487 g/mol) can be dissolved

in 500 mL of water at 25℃?

Ksp for Ba(IO3)2 is 1.57×10^(-9).

 

 

What mass in grams can be dissolved of barium iodate Ba(IO3)2

in 500 mL of distilled deionized water?

Ksp of Ba(IO3)2 is 1.57×10^(-9).

 

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Ba(IO3)2(s) ⇌ Ba^2+(aq) + 2IO3^-(aq)

Ksp = [Ba^2+] [IO3^-]^2 = (s) (2s)^2 = 4s^3

( 참고 https://ywpop.tistory.com/8434 )

 

 

 

s = [Ksp / 4]^(1/3)

= [(1.57×10^(-9)) / 4]^(1/3)

= 0.000732 M

---> Ba(IO3)2의 몰용해도

 

 

 

(0.000732 mol/L) (0.500 L) (487 g/mol)

= (0.000732) (0.500) (487)

= 0.178 g

 

 

 

답: 0.178 g

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/20844 ] 0.0200 M Ba(NO3)2 용액에서 Ba(IO3)2의 몰용해도

 

 

 

[키워드] Ba(IO3)2의 몰용해도 기준문서