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kozerog [31]
3 years ago
15

What is the molarity of a naoh solution if 39.1 ml of a 0.112 m h2so4 solution is required to neutralize a 25.0-ml sample of the

naoh solution?
Chemistry
1 answer:
aleksandrvk [35]3 years ago
8 0
The reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry  of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL  - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M

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Answer:

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Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 55 L

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Final volume (V₂) =?

The final volume of the gas can be obtained as follow:

P₁V₁ / T₁ = P₂V₂ / T₂

3.2 × 55 / 520 = 8.4 × V₂ / 760

176 / 520 = 8.4 × V₂ / 760

Cross multiply

520 × 8.4 × V₂ = 176 × 760

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Divide both side by 4368

V₂ = 133760 / 4368

V₂ = 30.62 L

Therefore, the new volume of the gas is 30.62 L

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