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>
Answer:
The correct answer is 2.016 x 10⁻¹⁷
Explanation:
We have the following chemical reactions and their equilibrium constants (K):
(1) H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq) K₁= 4.20×10⁻⁷
(2) HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
And we have to obtain K for the following reaction:
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)
If we add equations (1) and (2) we obtain the the desired equation. Remember that when we add chemical equations, the global equilibrium constant is the product of the constants.
H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + <em><u>HCO₃⁻(aq)</u></em> K₁= 4.20×10⁻⁷
+
<em><u>HCO₃⁻(aq)</u></em> + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
-------------------------------------------------------------
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq) K= K₁ x K₂
K = K₁ x K₂ = (4.20×10⁻⁷) x (4.80×10⁻¹¹) = 2.016 x 10⁻¹⁷
1, 4, 1, 2, 1
Explanation:
We have the following chemical reaction:
MnO₂ + HCl → MnCl₂ + H₂O + Cl₂
To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.
So the balanced chemical equation is:
MnO₂ + 4 HCl → MnCl₂ + 2 H₂O + Cl₂
Learn more about:
balancing chemical equations
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