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A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL of 0.235 M H2SO4 will neutralize this solution?

Please!!

User Shaahiin
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1 Answer

7 votes

Answer:

2.809 L of H₂SO₄

Step-by-step explanation:

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

First: We need to write the balanced equation for the reaction.

  • The reaction between NaOH and H₂SO₄ is a neutralization reaction.
  • The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Second: We calculate the umber of moles of NaOH used

  • Number of moles = Mass ÷ Molar mass
  • Molar mass of NaOH is 40.0 g/mol
  • Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

= 0.33 moles

Third: Determine the number of moles of the acid, H₂SO₄

  • From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
  • Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
  • Thus, Moles of H₂SO₄ = moles of NaOH × 2

= 0.33 moles × 2

= 0.66 moles of H₂SO₄

Fourth: Determine the Volume of the acid, H₂SO₄ used

  • When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
  • That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

= 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

User Mo Hossny
by
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