HCN 용액의 이온화 백분율

HCN 용액의 이온화 백분율

 

 

(a) Find the percent ionization of a 0.125 M HCN aqueous solution. The Ka for HCN is 4.9×10^(-10).

 

(b) Calculate the percent ionization of 0.20 M solution of hydrocyanic acid, HCN. Ka for HCN = 4.9×10^(-10).

 

(c) 0.100 M 사이안화수소산(HCN)

 

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(a)

Ka = x^2 / 0.125 = 4.9×10^(-10)

( 식 설명 https://ywpop.tistory.com/4294 )

 

x = [4.9×10^(-10) × 0.125]^(1/2)

= 7.83×10^(-6) M = [H^+]

 

 

[7.83×10^(-6)] / 0.125 × 100 = 0.00626%

( 참고 https://ywpop.tistory.com/3146 )

 

 

 

(b)

Ka = x^2 / 0.2 = 4.9×10^(-10)

x = [4.9×10^(-10) × 0.2]^(1/2)

= 9.90×10^(-6) M = [H^+]

 

 

[9.90×10^(-6)] / 0.2 × 100 = 0.00495%

 

 

 

(c)

Ka = x^2 / 0.100 = 4.9×10^(-10)

x = [4.9×10^(-10) × 0.100]^(1/2)

= 7.00×10^(-6) M = [H^+]

 

 

[7.00×10^(-6)] / 0.100 × 100 = 0.00700%