Using Henderson-hasselbach equation, pH=pka+log
[base]/[acid]
Or, 5.00=4.74+log[acetate]/[acetic acid]
Or,0.26= log[acetate]/[acetic acid]
[acetate]/[acetic acid]=10^0.26=1.82
or, [acetate]/[acetic acid]=1.82
As total molarity of the buffer=0.1M=0.1 mol/L
[acetate]+[acetic acid]=0.1 mol/L
Or, [acetate]=1.82[acetic acid]=1.82
(M-[acetate]=0.182-1.82[acetate]
Or, [acetate]=0.182-1.82[acetate]
2.182 [acetate]=0.182M
[acetate]=0.182M/2.182=0.0834 M
Molarity of acetic acid=0.1-0.0834=0.0166M
As the total volume of buffer=2*102ml=204ml
Moles of acetic acid=0.0166M*204ml*10^-3 L/ml=3.386 *10^-3
mol
Moles of acetate=0.0834 mol/L*204*10^-3L=17.0136 *10^-3 mol
Now, as Moles ofHCl added=4.40ml*0.390M=4.40*10^-3 L*0.390
mol/L=1.716*10^-3 moles
So moles of acetate neutralized =1.716*10^-3 moles
New moles of acetate=17.0136 *10^-3 mol-1.716*1.716*10^-3
moles
=15.298**10^-3 moles
Acetate +HCl=acetic acid +cl-
New moles of acetic acid=3.386 *10^-3 mol+1.716*10^-3
moles=5.102*10^-3 moles
Using Henderson-hasselbach equation, pH=pka+log
[base]/[acid]
New pH=4.74+ log (15.298**10^-3 moles)/( 5.102*10^-3 moles
        Ph=4.74+log
2.998=4.74+0.477=5.22
pH=5.22
pH=5.00+0.22
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