0.10 M Na2CO3 50.0 mL 0.10 M Ba(NO3)2 20.0 mL BaCO3

0.10 M Na2CO3 50.0 mL 0.10 M Ba(NO3)2 20.0 mL BaCO3

 

 

0.10 M Na2CO3 50.0 mL에

0.10 M Ba(NO3)2 20.0 mL를 가하면

BaCO3가 침전될 것인가?

 

 

If 20.0 mL of 0.10 M Ba(NO3)2 are added to

50.0 mL of 0.10 M Na2CO3,

will BaCO3 precipitate?

The Ksp of BaCO3 is 8.1×10^(-9).

 

 

A sample of 20.0 mL of 0.10 M Ba(NO3)2

is added to 50.0 mL of 0.10 M Na2CO3.

You wish to determine whether BaCO3 will precipitate.

Ksp = 8.1×10^(-9)

 

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(0.10 mol/L) (20.0/1000 L) = 0.002 mol Ba(NO3)2

( 참고 https://ywpop.tistory.com/7787 )

 

0.002 mol / (70.0/1000 L) = 0.02857 M Ba(NO3)2

= 0.02857 M Ba^2+

 

 

 

(0.10 mol/L) (50.0/1000 L) = 0.005 mol Na2CO3

 

0.005 mol / (70.0/1000 L) = 0.07143 M Na2CO3

= 0.07143 M CO3^2-

 

 

 

Ba(NO3)2(aq) + Na2CO3(aq) → BaCO3(s) + 2NaNO3(aq)

 

 

 

Q = [Ba^2+] [CO3^2-]

= (0.02857) (0.07143)

= 0.002041

 

 

 

Q > Ksp 이므로,

BaCO3 침전이 생성된다.

( 참고 https://ywpop.tistory.com/12527 )

 

 

 

 

[키워드] BaCO3 침전 기준문서, Q와 Ksp 비교 기준문서, Ksp와 Q 비교 기준문서