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
( 참고 https://ywpop.blogspot.com/2024/01/feoh3-10-7-m-oh.html )
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
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