when 50 ml of 0.1 M NaOH is added to 50 ml of 0.05 M CH3COOH solution, what will be the pH of the final solution? given that pKa=pKb= 4.7447 and pKw= 14. please give me a detailed solution for this.

Dear Parth,

5.2 = 4.75 + log 0.05 / x

5.2 - 4.75 = 0.45

10^0.45 = 0.05/x

2.82 = 0.05 / x

x = concentration acetic acid = 0.0177 M

moles acetic acid = 1 L x 0.0177 = 0.0177
moles acetate = 1 L x 0.05 = 0.05

moles OH- added = 0.050 L x 0.1 M = 0.0050

CH3COOH + OH- = CH3COO- + H2O
moles acetic acid = 0.0177 - 0.0050=0.0127
moles acetate = 0.05 + 0.0050=0.055

pH = 4.75 + log 0.055/ 0.0127=5.39

Best Of luck

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