(a) Explanation The reaction between HNO 3 and KOH is given as, KOH ( a q ) + HNO 3 ( a q ) → KNO 3 ( a q ) + H 2 O ( l ) Conversion of 1 mL into L is done as, 1 mL = 10 − 3 L Therefore, conversion of 24.9 mL into L is, 24.9 mL = 24.9 × 10 − 3 L = 0.0249 L Therefore, conversion of 25.0 mL into is, 25.0 mL = 25.0 × 10 − 3 L = 0.025 L As the salt formed by the reaction between a strong acid and a strong base is completely neutral, it does not affect the pH of the solution. The initial number of moles of H + is calculated by the formula, n H + = M H + × V 1 Where, n H + is the moles of H + in HNO 3 solution. M H + is the concentration of HNO 3 solution. V 1 is the volume of HNO 3 solution. Substitute the values of M H + and V 1 in the above formula. n H + = 0.100 M × 0.0249 L = 2.49 × 10 − 3 mol The initial number of moles of OH − is calculated by the formula, n OH − = M OH − × V 2 Where, n OH − is the moles of OH − in KOH solution. M OH − is the concentration of KOH solution. V 2 is the volume of KOH solution. Substitute the values of M H + and V 1 in the above formula. n OH − = 0.100 M × 0.025 L = 2.5 × 10 − 3 mol The reaction between a strong acid and strong base is given as, H + ( a q ) + OH - ( a q ) → H 2 O ( l ) The stoichiometry of the above reaction is, H + ( a q ) + OH − ( a q ) → H 2 O ( l ) Initial ( mol ) : 2 (b) Interpretation Introduction To determine: The pH when 25.1 mL of 0.100 M HNO 3 has been added to 25.0 mL of 0.100 M KOH solution.

Chemistry: The Central Science, Books a la Carte Plus MasteringChemistry with eText -- Access Card Package (13th Edition)

13th Edition

ISBN: 9780321934826

Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus

Publisher: PEARSON

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Chapter 3.6, Problem 3.16.2PE

(a)

Interpretation Introduction

To determine: The pH when 24.9 mL of 0.100 M

HNO3 has been added to 25.0 mL of 0.100 M

KOH solution.

(b)

Interpretation Introduction

To determine: The pH when 25.1 mL of 0.100 M

HNO3 has been added to 25.0 mL of 0.100 M

KOH solution.

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Chapter 3.6, Problem 3.16.1PE

Chapter 3.6, Problem 3.17.1PE

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Chemistry: The Central Science, Books a la Carte Plus MasteringChemistry with eText -- Access Card Package (13th Edition)

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The pH when 24.9 mL of 0.100 M HNO 3 has been added to 25.0 mL of 0.100 M KOH solution.