Answered: 50.0 mL of 0.0500 M HONH2 (a weak base;…

50.0 mL of 0.0500 M HONH2 (a weak base; Kb = 1.10e-08) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added: (d) 18.8 mL pH = (e) 25.0 mL pH = (f) 42.5 mL pH =

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter14: Principles Of Neutralization Titrations

Section: Chapter Questions

Problem 14.38QAP

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Concept explainers

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.

Question

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50.0 mL of 0.0500 M HONH₂ (a weak base; K_b = 1.10e-08) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added:

(d) 18.8 mL
pH =

(e) 25.0 mL

pH =

(f) 42.5 mL

pH =

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Consider the titration of 50.0 mL of 0.0500 M HONH2 (a weak base; Kb = 1.10e-08) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mLpH = (b) 6.3 mLpH = (c) 12.5 mLpH = (d) 18.8 mLpH = (e) 25.0 mLpH = (f) 42.5 mLpH =