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2 years ago

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Chemistry

2 answers:

Hitman42 [59]2 years ago

8 0

Equilibrium constant for the reaction is

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

Let [Al³+] be s
[OH-] is 10^{-7}

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

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

liraira [26]2 years ago

6 0

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.

<h3>Equation</h3>

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

<h3>Solving for [Al₃⁺³]</h3>

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.

<em>Additional comment</em>

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.

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