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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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