Answer:
OH⁻ = 1.05×10⁻¹³ M
Explanation:
Given data;
H₃O⁺ = 9.61×10⁻² M
OH⁻ = ?
Solution:
First of all we will calculate the pH.
pH = -log [H⁺]
pH = -log [H₃O⁺]
Because pH is also the hydronium ion concentration.
pH = -log [ 9.61×10⁻²]
pH = 1.02
It is known that,
pH + pOH = 14
1.02 + pOH = 14
pOH = 14 - 1.02
pOH = 12.98
Concentration of OH⁻.
pOH = -log [OH⁻]
12.98 = -log [OH⁻]
OH⁻ = 10∧-12.98
OH⁻ = 1.05×10⁻¹³ M
MgCl2 = 1Mg + 2Cl = 1(24.3) + 2(35.45) = 95.2g/1mole
7.50moles MgCl2 x 95.2g MgCl2 = 714g MgCl2
net ionic equation:
SO₄²⁻ (aq) + Ba²⁺ (aq) → BaSO₄ (s)
Explanation:
We have the following chemical reaction:
H₂SO₄ (aq) + Ba(OH)₂ (aq) → BaSO₄ (s) + 2 H₂O (l)
Now we write the ionic equation:
2 H⁺ (aq) + SO₄²⁻ (aq) + Ba²⁺ (aq) + 2 OH⁻ (aq) → BaSO₄ (s) + 2 H⁺ + 2 OH⁻ (aq)
We remove the spectator ions to obtain the net ionic equation:
SO₄²⁻ (aq) + Ba²⁺ (aq) → BaSO₄ (s)
where we have:
(aq) - aqueous
(s) - solid
(l) - liquid
Learn more about:
net ionic equation
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