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.
1 answer:
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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