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tester [92]
3 years ago
6

.950 L of .420 M H2SO4 is mixed with .900 L of .260 M KOH. What concentration of sulfuric acid remains after neutralization?

Chemistry
1 answer:
kipiarov [429]3 years ago
3 0
H₂SO₄:

V=0,95L
Cm=0,420mol/L

n = CmV = 0,42mol/L * 0,95L = 0,399mol

KOH:

V=0,9L
Cm=0,26mol/L

n = CmV = 0,26mol/L * 0,9L = 0,234mol

H₂SO₄            +           2KOH ⇒ K₂SO₄ + 2H₂O
1mol                :           2mol
0,399mol         :           0,234mol
                                    limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol

n = 0,282mol
V = 0,950L + 0,900L = 1,85L

Cm = n / V = 0,282mol / 1,85L ≈ 0,152M
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The equilibrium constant for the reversible reaction = 0.0164

Explanation:

At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.

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A ⇌ B

Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]

The rate of forward reaction = |r₁| = k₁ [A]

The rate of backward reaction = |r₂| = k₂ [B]

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k₁ = 0.010 s⁻¹

k₂ = 0.0610 s⁻¹

|r₁| = 0.010 [A]

|r₂| = 0.016 [B]

At equilibrium, the rate of forward and backward reactions are equal

|r₁| = |r₂|

k₁ [A] = k₂ [B] (eqn 1)

Note that equilibrium constant, K, is given as

K = [B]/[A]

So, from eqn 1

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[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164

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