Explanation Given, Titration of 50 .00 mL of 0 .250 M HNO 3 with 0 .500 M KOH after addition of 0, 12 .50, 25 .00, and 40 .00 mL KOH Addition of 0 mL KOH : Given system is titration of 50 .00 mL of 0 .250 M HNO 3 with 0 .500 M KOH after addition of 0 mL KOH The concentration of H N O 3 0.25 M, pH = - log ( H 3 O + ) pH = - log ( 0.25 ) pH = 0 .60 Addition of 12 .50 mL KOH : Given system is titration of 50 .00 mL of 0 .250 M HNO 3 with 0 .500 M KOH after addition of 12 .50 mL KOH The given titration is shown below, HNO 3 (aq) + KOH (aq) → KNO 3 (aq) + H 2 O (l) Moles of H N O 3 is calculated as follows, Moles of HNO 3 = 0 .05 L of HNO 3 × 0 .250 mol HNO 3 1 L of HNO 3 Moles of HNO 3 = 0 .0125 moles Moles of KOH is calculated as follows, Moles of NaOH = 0 .0125 L of NaOH × 0 .5 mol NaOH 1 L of NaOH Moles of NaOH = 6 .25 × 10 − 3 moles ICE table: HNO 3 (aq) + KOH (aq) → KNO 3 (aq) + H 2 O (l) S (mol) 0 .0125 moles 6 .25 × 10 − 3 moles 0 Excess R (mol) -6 .25 × 10 − 3 moles -6 .25 × 10 − 3 moles 6 .25 × 10 − 3 moles 6 .25 × 10 − 3 moles F(mol) 6 .25 × 10 − 3 moles 0 6 .25 × 10 − 3 moles Excess The total volume of the given solution is 50 ml + 12.5 ml = 62.5 ml (or) 0.0625 L C o n c e n t r a t i o n o f a c i d = M o l e s o f e x c e s s a c i d T o t a l v o l u m e C o n c e n t r a t i o n o f a c i d = 6.25 × 10 − 3 M 0.0625 L C o n c e n t r a t i o n o f a c i d = 0 .106 M The concentration of H N O 3 0.08 M, pH = - log ( H 3 O + ) pH = - log ( 0.1 ) pH = 1 .0 Addition of 25 .00 mL KOH : Given system is titration of 50 .00 mL of 0 .250 M HNO 3 with 0 .500 M KOH after addition of 25 .00 mL KOH The given titration is shown below, HNO 3 (aq) + KOH (aq) → KNO 3 (aq) + H 2 O (l) Moles of H N O 3 is calculated as follows, Moles of HNO 3 = 0 .05 L of HNO 3 × 0 .250 mol HNO 3 1 L of HNO 3 Moles of HNO 3 = 0 .0125 moles Moles of KOH is calculated as follows, Moles of NaOH = 0

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Author: David Reger; Scott Ball; Daniel Goode

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

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Chapter 16, Problem 16.20QE

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Interpretation:

The pH for the titration of 50.00 mL of 0.250 M HNO3 with 0.500 M KOH after addition of 0, 12.50, 25.00, and 40.00 mL KOH has to be calculated and also the titration curve has to be sketched.

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Chapter 16, Problem 16.19QE

Chapter 16, Problem 16.21QE

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Chapter 16 Solutions

Chemistry Principles And Practice

Ch. 16 - Prob. 16.1QE Ch. 16 - Sketch a titration curve for the titration of...Ch. 16 - Prob. 16.4QE Ch. 16 - Prob. 16.5QE Ch. 16 - Explain why the HendersonHasselbalch equation...Ch. 16 - Prob. 16.7QE Ch. 16 - Prob. 16.8QE Ch. 16 - Prob. 16.9QE Ch. 16 - Prob. 16.11QE Ch. 16 - Prob. 16.13QE

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Acid-Base Titration | Acids, Bases & Alkalis | Chemistry | FuseSchool; Author: FuseSchool - Global Education;https://www.youtube.com/watch?v=yFqx6_Y6c2M;License: Standard YouTube License, CC-BY

The pH for the titration of 50 .00 mL of 0 .250 M HNO 3 with 0 .500 M KOH after addition of 0, 12 .50, 25 .00, and 40 .00 mL KOH has to be calculated and also the titration curve has to be sketched.

Solution Summary: The author explains that the pH for the titration of 50 .00 mL has to be calculated and sketched.

Concept explainers

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Interpretation:

The pH for the titration of 50.00 mL of 0.250 M HNO3 with 0.500 M KOH after addition of 0, 12.50, 25.00, and 40.00 mL KOH has to be calculated and also the titration curve has to be sketched.

Want to see the full answer?

Chapter 16 Solutions

The pH for the titration of 50 .00 mL of 0 .250 M HNO 3 with 0 .500 M KOH after addition of 0, 12 .50, 25 .00, and 40 .00 mL KOH has to be calculated and also the titration curve has to be sketched.

Solution Summary: The author explains that the pH for the titration of 50 .00 mL has to be calculated and sketched.

Concept explainers

Videos

Interpretation: The pH for the titration of 50.00 mL of 0.250 M HNO3 with 0.500 M KOH after addition of 0, 12.50, 25.00, and 40.00 mL KOH has to be calculated and also the titration curve has to be sketched.

Want to see the full answer?

Chapter 16 Solutions

Interpretation:

The pH for the titration of 50.00 mL of 0.250 M HNO3 with 0.500 M KOH after addition of 0, 12.50, 25.00, and 40.00 mL KOH has to be calculated and also the titration curve has to be sketched.