Fe(OH)3 Ksp 4×10^(-38) mol/L water pH 5.0 buffer pH 11.0 buffer

Calculate the solubility (in mol/L) of Fe(OH)3 (Ksp = 4×10^(-38))

in each of the following situations:

a) Pure water (assume that the pH is 7.0 and remains constant)

b) A solution buffered at pH = 5.0

c) A solution buffered at pH = 11.0

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Fe(OH)3(s) ⇌ Fe^3+(aq) + 3OH^-(aq)

Ksp = [Fe^3+] [OH^-]^3

( 참고 https://ywpop.tistory.com/8434 )

a) Pure water (assume that the pH is 7.0 and remains constant)

pOH = 14.00 – pH

= 14.00 – 7.0 = 7.0

---> [OH^-] = 10^(-7.0) M

Ksp = [Fe^3+] [OH^-]^3

[Fe^3+] = Ksp / [OH^-]^3

= (4×10^(-38)) / (10^(-7.0))^3

= 4×10^(-17) M

Fe(OH)3 : Fe^3+ = 1 : 1 계수비(= 몰수비) 이므로,

[Fe^3+] = [Fe(OH)3] = 4×10^(-17) M

b) A solution buffered at pH = 5.0

pOH = 14.00 – 5.0 = 9.0

---> [OH^-] = 10^(-9.0) M

( ∵ 완충용액이므로, [OH^-] = 10^(-9.0) M )

[Fe^3+] = Ksp / [OH^-]^3

= (4×10^(-38)) / (10^(-9.0))^3

= 4×10^(-11) M

c) A solution buffered at pH = 11.0

pOH = 14.00 – 11.0 = 3.0

---> [OH^-] = 10^(-3.0) M

[Fe^3+] = Ksp / [OH^-]^3

= (4×10^(-38)) / (10^(-3.0))^3

= 4×10^(-29) M

답:

a) 4×10^(-17) M

b) 4×10^(-11) M

c) 4×10^(-29) M

[키워드] Fe(OH)3의 몰용해도 기준, Fe(OH)3의 Ksp 기준, Fe(OH)3의 몰용해도 사전, Fe(OH)3의 Ksp 사전