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20 mL of 0 . 1 MNaOH is added to 50 mL of 0 . 1 M acetic acid solution. The pH of the resulting solution is × 10 - 2 . (Nearest integer) Given : pK a CH 3 COOH = 4 . 76 log 2 = 0 . 30 log 3 = 0 . 48

Correct answer

0

Step-by-step solution

When a strong base is added to a weak acid solution, it results in the formation of a salt. Here, acid is present in a limiting reagent and base is present in excess amounts. So, by using the pH formula: pH = pKa + log salt acid CH 3 COOH + NaOH ⟶ CH 3 COONa + H 2 O 5                                 2                                 —   &#1

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