Using a 0.30 M phosphate buffer with a pH of 6.5, you add 0.80 mL of 0.55 M NaOH to 41 mL of the buffer. What is the new pH of the solution?
As the pH is around the pka2 value (7.21) , you have the species H2PO4- and HPO42- (mainly).
First you need to obtain the initial concentrations, these can be obtained with the Henderson-Hasselbalch equation:
that with the given values can be written as:
the other equation that we need is the total concentration:
isoltaing [HPO42- ] we have:
this value can be substituted in the Henderson-Hasselbalch equation to obtain [H2PO4-]:
Now we can know [HPO42- ]:
Next, we have to know the number of moles in the given volume of buffer:
We can calculate the final concentrations if we know the number of moles of base that will react, this is obtained as follows:
So, knowing that each mol of the buffer acid will react with one mole of added base to form one mol of buffer base:
In the Henderson-Hasselbalch equation we can write number of moles instead of concentrations because they are in the same volume:
Using a 0.30 M phosphate buffer with a pH of 6.5, you add 0.80 mL of...
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