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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

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BuyFindarrow_forward

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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86. What volume of 0.104 N H2SO4is required to neutralize

15.2 mL of 0.152 N NaOH? What volume of 0.104 M H2SO4is required to neutralize 15.2 mL of 0.152 M NaOH?

H 2 SO 4 ( aq ) + 2NaOH ( aq ) Na 2 SO 4 ( aq ) + 2H 2 O ( l )

Interpretation Introduction

Interpretation:

The volume of 0.104N

H2SO4 required to neutralize the given NaOH solution is to be calculated. The volume of 0.104M

H2SO4 required to neutralize the given NaOH solution is to be calculated.

Concept Introduction:

A chemical reaction that occurs between acid and base is known as neutralization reaction. In the neutralization reaction, the OH ions of the base combines with H+ ions of an acid. During neutralization reaction, water is released as a side product.

Molarity is used to determine the concentration of the substance in a solution.

Explanation

The value of N1, N2 and V1 is given to be 0.152N, 0.104N and 15.2mL respectively.

The relationship between normality and volume of both solutions is shown below.

Where,

  • N1 is the normality of NaOH solution.
  • V1 is the volume of NaOH solution.
  • N2 is the concentration of H2SO4 solution.
  • V2 is the volume of H2SO4 solution.

Rearrange an above expression to calculate V2 which is needed to neutralize the H2SO4 solution.

V2=N1×V1N2        (1)

Substitute the value of N1, N2 and V1 in the equation (1).

V2=0.152N×15.2mL0.104N=22.2mL

Therefore, the volume of 0.104N

H2SO4 required to neutralize the given NaOH solution is 22.2mL.

The molarity of H2SO4 solution is given to be 0.104M. One equivalent of sulfuric acid H2SO4 produces two moles of H+ ions. One equivalent of sodium hydroxide NaOH produces one mole of OH ions. Hence, the molarity of NaOH is equal to normality of NaOH.

The normality of H2SO4 solution is calculated by the formula,

Normality=Molarity×NumberofH+ions        (2)

Substitute the values of molarity and number of H+ ions in the equation (2)

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