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>
A the nucleus emits partials and or energy
Answer:
0.065M
Explanation:
The reaction expression is given as:
HCl + KOH → KCl + H₂O
Given parameters:
Volume of KOH = 25.81mL = 0.02581L
Molarity of KOH = 0.125M
Volume of HCl = 50mL = 0.05L
Unknown:
Molarity of the acid = ?
Solution:
To find the molarity of the acid, HCl, let us use the mole concept.
First, find the number of moles of the base, KOH;
Number of moles of KOH = molarity of KOH x volume of KOH
Number of moles of KOH = 0.125 x 0.0258 = 0.003225mol
From the balanced reaction equation:
1 mole of KOH reacted with 1 mole of HCl
0.003225mol of KOH will react with 0.003255mole of HCl
Molarity of HCl =
Molarity of HCl = = 0.065M
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