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500 mL of 0.2 M aqueous solution of acetic acid is mixed with 500 mL of 0.2 M HCl at 25^(@)C. (i) Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution. (ii) If 6 g of NaOH is added to the above solution, determine the final pH [Assume there is no change in volume on mixing : K_(a)of acetic acid is 1.75xx10^(-5) " mol " L^(-1)] |
| Answer» <html><body><p></p>Solution :(i) Millimoles of `CH_(3)CO OH = <a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a> xx 0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> = 100` <br/> Millimoles of HCl `= 500xx 0.2=100` <br/> Final volume after mixing `= 500 + 500 = 1000 ` mL <br/> `:. [ CH_(3)CO OH]=(100)/(1000) = 0.1 M,"" [HCl ] = (100)/(1000) = 0.1 M` <br/> `{:(,CH_(3)CO OH ,hArr,CH_(3)CO O^(-),+,H^(+),,),("Before dissociation",0.1 M ,," "0,,0.1 M,,),(,,,,,("from HCl"),,),("After dissociation",(0.1-<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>),," "x,,(0.1+x),,):}` <br/> `K_(a) = (x(0.1+))/((0.1+x))` <br/> As in the presence of HCl, dissociation of `CH_(3)CO OH` will be very small (due to common ion effect ),x is very very small. Hence, <br/> `K_(a) = (x(0.1))/(0.1) = x = 1.75xx10^(-5) `<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> `L^(-1) ` (Given) <br/> `:.` Degree of dissociation `= (x)/(0.1) = (1.75xx10^(-5))/(0.1) = 1.75xx10^(-4)= 0.00175 %` <br/> Further, `[H^(+)]= 0.1 + x ~~ 0.1 :. pH = - log 0.1 = 1 ` <br/> (ii) 6g of <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> = `(6)/(40) ` mole = 0.15 mole <br/> Hence, now the equilibrium will be<br/> `{:(,CH_(3)CO OH ,+,HCl,+,NaOH,hArr,CH_(3)CO ONa ,+,NaCl,+,H_(2)O),("Initial",0.1,,0.1,,0.15,,0,,0,,0),("At eqm.",0.05,,0,,0,,0.05,,0,,0):}` <br/> Thus, the solution will now be 0.05 M in `CH_(3)CO OH` and 0.05 M in `CH_(3)CO ONa`, i.e., it is acidic buffer. <br/> `pH = - log K_(a) + log. (["Salt"])/(["Acid"]) = - log (1.75xx10^(-5)) + log (0.05)/(0.05) = 4. 757`</body></html> | |