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JEE Main20207 Jan 2020Evening ShiftChemistryIonic EquilibriumActual

3 g of acetic acid is added to 250 m L of 0.1 M H C l and the solution made up to 500 m L . To 20 m L of this solution 1 2 m L of 5 M N a O H is added. The p H of the solution is ________ [Given: p K a of acetic acid = 4.75 , molar mass of acetic acid 60 g / m o l , log ⁡ 3 = 0.4771 , Neglect any changes in volume]

Correct answer

0

Step-by-step solution

millimole of acetic acid in 20 m L = 2 millimole of H C l in 20 m L = 1 millimole of N a O H = 2.5 p H = P K a + log ⁡ 3 / 2 2 = 4.74 + log ⁡ 3 = 4.74 + 0.48 = 5.22

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