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Vikentia [17]
3 years ago
6

what is the concentration of an NaOH solution that requires 50 mL of a 1.25 M H2SO4 solution to neutralize 78.0 ml of NaOH​

Chemistry
1 answer:
Dimas [21]3 years ago
3 0

Answer:

  • The answer is the concentration of an NaOH = 1.6 M

Explanation:

The most common way to solve this kind of problem is to use the formula  

  • C₁ * V₁ = C₂ * V₂

In your problem,

For NaOH

C₁ =??     v₁= 78.0 mL = 0.078 L

For H₂SO₄

C₁ =1.25 M     v₁= 50.0 mL = 0.05 L

but you must note that for the reaction of NaOH with H₂SO₄

2 mol of NaOH raect with 1 mol H₂SO₄

So, by applying in above formula

  • C₁ * V₁ = 2 * C₂ * V₂
  • (C₁ * 0.078 L) = (2*  1.25 M * 0.05 L)
  • C₁ = (2*  1.25 M * 0.05 L) / (0.078 L) = 1.6 M  

<u>So, the answer is the concentration of an NaOH = 1.6 M</u>

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\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

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\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]

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Putting values in above equation, we get:

\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ

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