What mass, in grams, of sodium bicarbonate, nahco3, is required to neutralize 1000.0 l of 0.350 m h2so4?
1 answer:
Hello!
The net chemical equation between Sodium Bicarbonate and H₂SO₄ is the following one:
2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂
To calculate the amount of Sodium Bicarbonate needed to neutralize 1000 L of 0,350 M H₂SO₄ we'll need to use the following conversion factor, to go from the volume of H₂SO₄ to grams of Sodium Bicarbonate:
So, to neutralize 1000 L of 0,350 moles of H₂SO₄ you'll need 58804,9 grams of NaHCO₃
The net chemical equation between Sodium Bicarbonate and H₂SO₄ is the following one:
2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂
To calculate the amount of Sodium Bicarbonate needed to neutralize 1000 L of 0,350 M H₂SO₄ we'll need to use the following conversion factor, to go from the volume of H₂SO₄ to grams of Sodium Bicarbonate:
So, to neutralize 1000 L of 0,350 moles of H₂SO₄ you'll need 58804,9 grams of NaHCO₃
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Answer:
0.3023 M
Explanation:
Let Picric acid =
So, +
⇄
+
The ICE table can be given as:
+
⇄
+
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant () = 0.42
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
=
=
= OR
= OR
= OR
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [ ] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).