이염기성 화합물 B. Kb1 = 1.00×10^(-5) 1.00×10^(-9)

이염기성 화합물 B. Kb1 = 1.00×10^(-5) 1.00×10^(-9)

 

 

The dibasic compound B forms BH^+ and BH2^2+

with Kb1 = 1.00×10^(-5) and 1.00×10^(-9).

Find the pH and concentrations of B, BH^+, and BH2^2+

in each of the following solutions:

a) 0.100 M B

b) 0.100 M BH^+Br^-

c) 0.100 M BH2^2+(Br^-)2.

For (b) use the approximation [BH^+] ≒ 0.100 M.

 

 

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a) 0.100 M B

 

B(aq) + H2O(l) ⇌ BH^+(aq) + OH^-(aq)

Kb1 = [BH^+] [OH^-] / [B]

1.00×10^(-5) = (x) (x) / (0.100-x)

( 참고 https://ywpop.tistory.com/4294 )

 

 

 

0.100-x ≒ 0.100 이라 근사처리하면,

x = [ (1.00×10^(-5)) × 0.100 ]^(1/2)

= 0.00100 M = [OH^-] = [BH^+]

 

 

 

[B] = 0.100 – 0.00100 = 0.099 M

 

 

 

pOH = -log[OH^-]

= -log(0.00100)

= 3.00

 

 

 

pH = 14.00 – 3.00 = 11.00

 

 

 

BH^+(aq) + H2O(l) ⇌ BH2^2+(aq) + OH^-(aq)

Kb2 = [BH2^2+] [OH^-] / [BH^+]

1.00×10^(-9) = (x) (0.00100+x) / (0.00100-x)

 

 

 

0.00100-x ≒ 0.00100

0.00100+x ≒ 0.00100

이라 근사처리하면,

x = 1.00×10^(-9) M = [BH2^2+]

 

 

 

 

a) 11.00, 0.0990 M, 9.95×10^-4 M, 1.00×10^-9 M

b) 7.00, 1.0×10^-3 M, 0.100 M, 1.0×10^-3 M

c) 3.00, 1.00×10^-9 M, 9.95×10^-4 M, 0.0990 M

 

 

 

 

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