Question

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Answer 1

Equilibrium constant for the reaction is

$\sf K_{sp}=[Al^{3+}][OH^-]^3$

• Let [Al³+] be s
• [OH-] is 10^{-7}

So

$\\ \rm\Rrightarrow s=\dfrac{3.0\times 10^{-34}}{(10^{-7})^3}[/tex[[tex]\\ \rm\Rrightarrow s=\dfrac{3.0\times 10^{-34}}{10^{-21}}$

$\\ \rm\Rrightarrow s=3.0\times 10^{-13}M$

Answer 2

Answer:

  3.0×10⁻¹³ M

Explanation:

The solubility product Ksp is the product of the concentrations of the ions involved. This relation can be used to find the solubility of interest.

Equation

The power of each concentration in the equation for Ksp is the coefficient of the species in the balanced equation.

  Ksp = [Al₃⁺³]×[OH⁻]³

Solving for [Al₃⁺³]

The initial concentration [OH⁻] is that in water, 10⁻⁷ M. The reaction equation tells us there are 3 OH ions for each Al₃ ion. If x is the concentration [Al₃⁺³], then the reaction increases the concentration [OH⁻] by 3x.

This means the solubility product equation is ...

  Ksp = x(10⁻⁷ +3x)³

For the given Ksp = 3×10⁻³⁴, we can estimate the value of x will be less than 10⁻⁸. This means the sum will be dominated by the 10⁻⁷ term, and we can figure x from ...

  3.0×10⁻³⁴ = x(10⁻⁷)³

Then x = [Al₃⁺³] will be ...

  $[\text{Al}_3^{\,+3}]=\dfrac{3.0\times10^{-34}}{10^{-21}}\approx \boxed{3.0\times10^{-13}\qquad\text{moles per liter}}$

We note this value is significantly less than 10⁻⁷, so our assumption that it could be neglected in the original Ksp equation is substantiated.

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Additional comment

The attachment shows the solution of the 4th-degree Ksp equation in x. The only positive real root (on the bottom line) rounds to 3.0×10^-13.