Part a a solution is made by mixing 9.00 mmol (millimoles) of ha and 3.00 mmol of the strong base. what is the resulting ph? express the ph numerically to two decimal places.

Assuming that the reaction is unimolecular on both reactant and product sides such that:

HA  +  OH-   -->  A-  +  H2O

Therefore the amount of A- left is:

mmol HA = 9.00 - 3.00 = 6.00 
mmol A- = 3.00 

pKa = - log Ka = 5.25 

Calculating for pH:
pH = 5.25 + log 3.00/6.00

well in order to answer this question you have to provide more information < 3

b is the correct answer

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