25.00 cm3 of propanoic acid, with a concentration of 0.120 mol dm−3, was pipetted into a conical flask. This solution was titrated with sodium hydroxide of concentration 0.150 mol dm−3. CH3CH2COOH + NaOH o CH3CH2COONa + H2O
(iii) Calculate the minimum volume of sodium hydroxide required to react with all of the propanoic acid. (2)
ans=20cm^3
iv) Calculate the pH when 40 cm3
of sodium hydroxide (an excess) was added.
ans:-(Excess NaOH = 20.0 cm3 in total volume of 65
cm3 ) [OH− ] = (20.0 x 0.15)/65 = 0.046154 [H+ ] = 1 x 10^−14/0.46154 = 2.1667 x 10−13 pH = (−log 2.1667 x 10−13 ) = 12.6642/12.7
can anyone explain why we consider the unreacted naoh =20 cm3 in the calculation, wont the reacted 20cm3 affect the ph as well?