Mg(OH)2의 몰용해도. pH 9.00 완충 용액에서

Mg(OH)2의 몰용해도. pH 9.00 완충 용액에서

 

 

What is the molar solubility of Mg(OH)2

in a buffer solution that has a pH of 9.00?

Mg(OH)2(s) ⇌ Mg^2+(aq) + 2OH^-(aq) ... Ksp = 1.8×10^(-11)

 

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Mg(OH)2(s) ⇌ Mg^2+(aq) + 2OH^-(aq)

Ksp = [Mg^2+] [OH^-]^2

 

 

 

pOH = 14.00 – pH

= 14.00 – 9.00

= 5.00

 

 

 

pOH = -log[OH^-] = 5.00

[OH^-] = 10^(-5.00) M

 

 

 

[Mg^2+] = Ksp / [OH^-]^2

= (1.8×10^(-11)) / (10^(-5.00))^2

= 0.18 M

 

 

 

답: 0.18 M