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Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

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Chapter
Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4, Problem 112GQ
Textbook Problem
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A solution of hydrochloric acid has a volume of 250. mL and a pH of 1.92. Exactly 250. mL of 0.0105 M NaOH is added. What is the pH of the resulting solution?

Interpretation Introduction

Interpretation:

 pH of the solution obtained by adding given volume of HCl with given volume of NaOH under given pH condition has to be determined.

Concept introduction:

  • Strong acids dissociates completely into ions in solution but weak acids do not.
  • pH of a solution is the negative of the base -10 logarithm of the hydronium ion concentration.

pH=-log[H3O+]

  • Concentration of hydronium ion [H3O+]=10-pH
  • For an acidic solution pH<7 and for a basic solution pH>7.
  • Amountofsubstance=Concnetrationofthesubstance×Volume
  • Concnetrationofthesubstance=AmountofsubstanceVolume

Explanation of Solution

In the given solution of HCl has a volume of 250.mL and a pH of 1.92

From its pH value Concentration of hydronium ion can be determined using the formula,

Concentration of hydronium ion [H3O+]=10-pH

The pH of the given solution is 1.92. Substituting this value in the above equation,

Concentration of hydronium ion is,

 [H3O+]=10-1.92 =0.012M

Amount of HCl in 250.mL of 0.012M solution can be calculated as follows,

AmountofHCl =ConcnetrationHCl×VolumeHCl =0.012molL×0.25L =0.003molHCl

Similarly the amount of NaOH in 250.mL of 0.0105M solution can be calculated.

AmountofNaOH =ConcnetrationNaOH×VolumeNaOH =0

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Chapter 4 Solutions

Chemistry & Chemical Reactivity
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