Part A – Calculate the molar solubility in water Mg(OH)2 is a sparingly soluble compound, in this case a base, with a solubility prod

Question

Part A – Calculate the molar solubility in water
Mg(OH)2 is a sparingly soluble compound, in this case a base, with a solubility product, Ksp, of 5.61×10 −11 . It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams.
Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in pure H2O?
Part B – Calculate the molar solubility in NaOH
Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in 0.160 M NaOH?

in progress 0
2022-05-11T19:19:06+00:00 1 Answer 0

Answers ( 1 )

  1. Jasmine
    0
    2022-05-11T19:20:38+00:00

    Part 1)

    when the balanced equation for this reaction is:

    and by using ICE table

    Mg(OH)2(s)u2194 Mg2+(aq) + 2OH-(aq)

    initial 0 0

    change +X +2X

    Equ X 2X

    When KSp expression = [Mg2+][OH-]

    when we have KSp = 5.61 x 10^-11

    and when we assumed [Mg2+] = X

    and [ OH-] = (2X)^2

    when we assume X is the value of molar solubility of Mg(OH)2

    so, by substitution:

    5.61 x 10^-11 = 4X^3

    u2234 X = 2.4 x 10^-4 M

    u2234 molar solubility of Mg(OH)2 = X = 2.4 x 10^-4 M

    Part 2)

    the molar solubility of Mg(OH)2 in 0.16 m NaOH we assumed it = X

    by using the ICE table:

    Mg(OH)2(s)u2192 Mg2+(aq) + 2OH-

    initial 0 0.16m

    change +X +2X

    equ X (0.16+2x)

    when Ksp = [mg2+][OH-]^2

    5.61 x 10^-11 = X * (0.16+2X)^2 by solving for X

    u2234 X = 1.3 x 10^-5 M

    u2234 the molar solubility = X = 1.3 x 10^-5 M

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