Write the equilibrium expression for the dissolution of Sr(OH)2 and calculate the pH of the resulting solution

Write the equilibrium expression for the dissolution of Sr(OH)2 and calculate the pH of the resulting solution when Ksp=1.23x10^-5 mol^3 L^-3??

Sr(OH)2 <--> Sr^2+ + 2 OH^-

Ksp = 1.23x10^-5 mol^3 L^-3

You know by equation:

[OH^-] = 2*[Sr^2+]

[Sr^2+] =0.5*[OH^-]

Ksp= [Sr^2+]*[OH^-]^2 = 0.5*[OH^-]*[OH^-]^2 = 0.5*[OH^-]^3

1.23*10^-5 = 0.5*[OH^-]^3

[OH^-]^3 = (1.23*10^-5/0.5)

[OH^-]^3 = 2.46*10^-5

[OH^-] = cuberoot 2.46*10^-5 = 0.0291 = 2.91*10^-2

[OH^-] = 2.91*10^-2

pH + pOH = 14

pH = 14 – pOH

Calculate pOH

pOH = - log [OH^-] = - log 2.91*10^-2 = - [log 2.91 + log 10^-2] = - [0.464 -2] = - (-1.536) = 1.536

pOH = 1.536

pH = 14 – pOH

pH = 14 – 1.536 = 12.464

Answer:

pH = 12.464

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