Question

Mr. banners working at the water treatment plant in his learning about coagulant aluminum hydroxide. He needs to determine how much aluminum ion can be present in water at pH = 6.5 before precipitation will occur. Find the aluminum ion concentration in milligrams per liter.

Answer 1

Given that pH = 6.5

so, pOH = 14-6.5 = 7.5

-log[OH^-] = 7.5

log[OH-] = -7.5

[OH-] = 10^-7.5

[OH-] = 3.162 x 10^-8 M

Let S be the solubility of Al(OH)3

Al(OH)₃ -----------> Al³⁺ + 3(OH^-)

[Al3+] = S g/L and [OH-] = 3S g/L

But [OH-] = 3.162 x 10^-8 M

So, 3S = 3.162 x 10^-8

S = 3.162 x 10^-8 / 3

S= 1.05 x 10^-8.

Hence concentration of [Al3+] = S = 1.05 x 10^-8 g/L

= 1.05 x 10^-8 x 10³ mg/L........(since, 1 g =10³ mg)

= 1.05 x 10^-5 mg/L

So, Aluminium ion concentration at pH 6.5 will be 1.05 x 10^-5 mg/L

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