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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.

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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

pH = 4.95 

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