now when the base is added, first a buffer solution is formed
until the point the whole of the acid is used up. then the ph rises
due to excess of naoh
so a)when 10ml is added,
milli moles of acid left=25*0.115-0.144*10
milli moles of salt formed=10*0.144
b)when 20 ml is added,
excess milli moles of the base=20*0.144-25*0.115
so pH=14+ log(0.005/(20+25))
c)when 30 ml is added,
excess milli moles of the base=30*0.144-25*0.115
so pH=14+ log(1.445/(30+25))
give me good review
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