Problem: Consider the titration of 100.0 mL of 0.200 M HONH 2 by 0.100 M HCl. (K b for HONH2 = 1.1 x 10-8.)a. Calculate the pH after 0.0 mL of HCl has been added.

FREE Expert Solution

We are asked to calculate the pH after 0.0 mL of HCl has been added.


Since no acid is added yet, we treat is as a simple weak base. 


Step 1. Set up ICE chart: 


Step 2. Calculate OH- 


Kb = [HONH3+][OH-][HONH2]1.1 × 10-8= [x][x][0.200 - x]1.1 × 10-8= x2[0.200 - x]

SInce Kb is small, assume x is negligible in the denominator. 

1.1 × 10-8= x2[0.200 ]x2 = (1.1 × 10-8)(0.200 )x = (1.1 × 10-8)(0.200 )

x = [OH-] = 4.69 x 10-5 M


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Problem Details

Consider the titration of 100.0 mL of 0.200 M HONH 2 by 0.100 M HCl. (K b for HONH2 = 1.1 x 10-8.)

a. Calculate the pH after 0.0 mL of HCl has been added.

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Weak Base Strong Acid Titrations concept. You can view video lessons to learn Weak Base Strong Acid Titrations. Or if you need more Weak Base Strong Acid Titrations practice, you can also practice Weak Base Strong Acid Titrations practice problems.

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.