Ionic EquilibriumMedium
Question
A definite volume of a N/20 CH3COOH (pKa = 4.7447) is titrated with a strong base (NaOH). It is found that 80 equal sized drops of NaOH, added from a burette effect the complete neutralisation. Find the pH, when the acid solution is neutralised to the extent of 20%.
Options
A.4.14
B.9.86
C.5.34
D.8.68
Solution
WA(CH3COOH) is titrated with SB(NaOH) and 20% of the WA is neutralised, so 20% of the salt of WA/SB (CH3COOH) is formed and 80% of the WA(CH3COOH) is left. So an acidic buffer is formed.
∴ [Acid] = 80; [Salt] = 20
pH = pKa + log
= 4.7447 + log
= 4.7447 + log
= 4.7447 - 2 log 2 = 4.7447 - 2 × 0.3010 = 4.14
∴ [Acid] = 80; [Salt] = 20
pH = pKa + log
= 4.7447 + log
= 4.7447 + log
= 4.7447 - 2 log 2 = 4.7447 - 2 × 0.3010 = 4.14
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
The molar solubility in terms of solubility product Ksp is given by the relation...For which of the following solutions must we consider the ionisation of water when calculating the pH or pOH :-...The reverse process of neutralisation is:...The product of reaction of I- with MnO4- in alkaline medium :-...The pH of the solution obtanied by mixing 10 mL of 10-1 N HCI and 10 mL of 10-1 N NaOH is:...