Question

What is the ph after 0.195 mol of naoh is added to the buffer from part a? assume no volume change on the addition of the base?

Answer 1

[HA] = 0.708/ 2.00 = 0.354 M 
[A-] = 0.609/2.00 =0.305 M 
pKa = 6.25 
pH = 6.25 + log 0.305 / 0.354 =6.19 

A- + H+ >> HA 
moles A- = 0.609 - 0.150 =0.459 
[A-]= 0.459/ 2.00 =0.230 M 
moles HA = 0.708 + 0.150 =0.858 
[HA] = 0.858/2.00 = 0.429 M 
pH = 6.25 + log 0.230/0.429 = 5.98 

HA + OH- >> A- + H2O 
moles HA = 0.708 - 0.195 =0.513 
[HA]= 0.513/2.00 =0.257 M 
moles A- = 0.609 + 0.195 =0.804 
[A-]= 0.804/2.00 = 0.402 M 

pH = 6.25 + log 0.402/0.257 =6.44


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