A) Show how to use the Ksp of anhydrous magnesium hydroxide (7. X 10^-12) to fin

Question

A) Show how to use the Ksp of anhydrous magnesium hydroxide (7. X 10^-12) to find the maximum [OH-] that is compatible with 0.1 M Mg(H2O)6^2+ at equalibrium. Calculate the pH. B) Show the calculation of the pH of a 1.0 M NH3/NH4 + buffer. When equal volumes of this buffer and 0.2 M Mg(H2O)6^2+ solution are mixed use your answer in part (a) to predict whether or not solid magnesium hydroxide would form. C) is this equalibrium, between the hydrated hydroxide and its ions attained rapidly? A) Show how to use the Ksp of anhydrous magnesium hydroxide (7. X 10^-12) to find the maximum [OH-] that is compatible with 0.1 M Mg(H2O)6^2+ at equalibrium. Calculate the pH. B) Show the calculation of the pH of a 1.0 M NH3/NH4 + buffer. When equal volumes of this buffer and 0.2 M Mg(H2O)6^2+ solution are mixed use your answer in part (a) to predict whether or not solid magnesium hydroxide would form. C) is this equalibrium, between the hydrated hydroxide and its ions attained rapidly? B) Show the calculation of the pH of a 1.0 M NH3/NH4 + buffer. When equal volumes of this buffer and 0.2 M Mg(H2O)6^2+ solution are mixed use your answer in part (a) to predict whether or not solid magnesium hydroxide would form. C) is this equalibrium, between the hydrated hydroxide and its ions attained rapidly?

Explanation / Answer

Mg(OH)2 Mg2+ + 2 OH¯

Ksp = [Mg2+] [OH¯]2

7 x 10-12= (s) (2s)2 = 4s3 (s is solubility)

s =( 7 x 10-12 / 4)1/3

s = 1.20 x 10-4

We have calculated is 's' and it is NOT the [OH-]. That value is '2s.'

[OH-] = 2.4 x 10-4

pOH = -log(2.4 x 10-4 ) = 3.62

pH = 14 - pOH

pH = 14 - 3.62 = 10.38

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