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


View Complete Written Solution
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.