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Determine the pH, pOH, [H+], and [OH−] of a solution in which 0.300 g of aluminum hydroxide is dissolved in 184 mL of solution.

Sagot :

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Answer:

[OH⁻] = 0.0627M

pOH = 1.20

pH = 12.8

[H⁺] = 1.59x10⁻¹³M

Explanation:

To solve this question we must, as first, find the molarity of Al(OH)₃ in the solution -Molar mass Al(OH)₃: 78.00g/mol-:

0.300g * (1mol/ 78.00g) = 3.846x10⁻³ moles

In 184mL = 0.184L:

3.846x10⁻³ moles / 0.184L = 0.0209M Al(OH)₃. Three times this molarity = [OH⁻]:

[OH⁻] = 0.0209M * 3

[OH⁻] = 0.0627M

pOH = -log [OH⁻] =

pOH = 1.20

pH = 14 - pOH

pH = 12.8

And [H⁺] = 10^-pH

[H⁺] = 1.59x10⁻¹³M

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