3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution $\frac{1}{2}$ mL of 5 M NaOH is added. The pH of the solution is?
[Given: pKa of acetic acid $= 4.75$ , molar mass of acetic acid $= 60 g / m o l, l o g 3 = 0.4771$ ] Neglect any changes in volume.

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mili mole of acetic acid in 20 ml = 2 mili mole of HCl in 20 ml = 1 mili mole of NaOH 20 ml = 2.5 N a O H ( l e f t ) 1.5 0 + C H 3 C O O H 2 0.5 → C H 3 C O O N a 0 1.5 + H 2 O 0 p H = p K a + l o g s a l t a c i d = 4.74 + l o g 1.5 0.5 = 4.74 + l o g 3 = 4.74 + 0.4771 = 5.22

Chemistry - Ionic Equilibrium Question with Solution | TestHub

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