To solve this problem, we need to understand the dissociation of Al(OH)3 in water. Al(OH)3 dissociates in water as follows: Al(OH)3 Al3+ + 3OH- This means that for every one mole of Al(OH)3 that dissociates, one mole of Al3+ and three moles of OH- are produced. The solubility product constant expression for this reaction is: Ksp = [Al3+][OH-]^3 Given that Ksp of Al(OH)3 at 25°C is 1×10^-32, we can set up the following equation: 1×10^-32 = [Al3+][OH-]^3 Since the concentration of Al3+ and OH- are the same, we can simplify this equation to: 1×10^-32 = [Al3+]^4 Taking the fourth root of both sides gives us the concentration of Al3+: [Al3+] = (1×10^-32)^(1/4) [Al3+] = 1×10^-8 M So, the concentration of the aluminum ion (Al3+) in pure water caused by the complete dissociation of Al(OH)3 is 1×10^-8 M.

Question

To solve this problem, we need to understand the dissociation of Al(OH)3 in water.

Al(OH)3 dissociates in water as follows:

Al(OH)3 <=> Al3+ + 3OH-

This means that for every one mole of Al(OH)3 that dissociates, one mole of Al3+ and three moles of OH- are produced.

The solubility product constant expression for this reaction is:

Ksp = [Al3+][OH-]^3

Given that Ksp of Al(OH)3 at 25°C is 1×10^-32, we can set up the following equation:

1×10^-32 = [Al3+][OH-]^3

Since the concentration of Al3+ and OH- are the same, we can simplify this equation to:

1×10^-32 = [Al3+]^4

Taking the fourth root of both sides gives us the concentration of Al3+:

[Al3+] = (1×10^-32)^(1/4)

[Al3+] = 1×10^-8 M

So, the concentration of the aluminum ion (Al3+) in pure water caused by the complete dissociation of Al(OH)3 is 1×10^-8 M.

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