Problem: Calculate the pH after 0.020 mole of HCl is added to 1.00 L of each of the following solutions:a. 0.100 M HONH2 (Kb = 1.1 x 10-8)b. 0.100 M HONH3Clc. pure H2Od. a mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl

🤓 Based on our data, we think this question is relevant for Professor Savizky's class at COLUMBIA.

FREE Expert Solution

For a.

mole HONH2=1.00 L(0.100 molL)=0.100 mol

Reaction: HONH2(aq) + HCl(aq) ⇌ HONH3+(aq) + H2O(l)

HONH3+(aq) + H2O(l) ⇌ HONH2(aq) + H3O+(aq)

Ka=KwKb=1.0×10-141.1×10-8=9.09×10-7


Ka=productsreactants=[HONH2][H3O+]HONH3+


Ka=(x)(x)0.020-x


initial HONH3+ concentrationKb=0.0209.09×107>>>x can be ignored in the denominator

Solve for x:

9.09×10-7=x20.020-xx2=(9.09×10-7)(0.020)

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Problem Details

Calculate the pH after 0.020 mole of HCl is added to 1.00 L of each of the following solutions:

a. 0.100 M HONH2 (Kb = 1.1 x 10-8)
b. 0.100 M HONH3Cl
c. pure H2O
d. a mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl

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Based on our data, we think this problem is relevant for Professor Savizky's class at COLUMBIA.

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