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vodomira [7]
3 years ago
15

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.

Chemistry
1 answer:
lys-0071 [83]3 years ago
3 0

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

<h3>[OH⁻] = 0.0627M</h3>

pOH = -log [OH⁻] =

<h3>pOH = 1.20</h3>

pH = 14 - pOH

<h3>pH = 12.8</h3>

And [H⁺] = 10^-pH

<h3>[H⁺] = 1.59x10⁻¹³M</h3>
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