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3g of acetic acid is added to 250 mL of 0.1M 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 acertic acid = 60 g/mol, log3 = 0.4771] Neglect any chnages in volume.
 

Answers (1)

best_answer

Given,

m mole of acetic acid in 20 mL = 2
m mole of HCl in 20 mL = 1
m mole of NaOH = 2.5

\mathrm{CH_{3}COOH\: +\: NaOH(remaining)\: \rightarrow \: CH_{3}COONa\: +\:water}
          2                              3/2                                      0                                 0
         0.5                             0                                       3/2            

\mathrm{pH\: =\: pK_{a}\: +\: log\frac{\frac{3}{2}}{2}}

pH = 4.74 + log 3

pH = 4.74 + 0.48 = 5.22

Thus, the correct answer is 5.22

Hence, the option number (1) is correct.

Posted by

Kuldeep Maurya

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