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3.0 g of pure acetic acid and 4.1 g of anhydrous sodium acetate are dissolved together in water and the solution is made up to 500 ml. Calculate the pH of the solution. Given K_(a) of acetic acid is 1.75xx10^(-5).

Answer» <html><body><p></p>Solution :Calculation of the concentration of acetic acid : <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a>/`<a href="https://interviewquestions.tuteehub.com/tag/dm-432223" style="font-weight:bold;" target="_blank" title="Click to know more about DM">DM</a>^(3)=Nxxg` eq. mass <br/> `"mass"//500cm^(3)=(Nxxg" eq. mass")/2rArrN_(CH_(3)<a href="https://interviewquestions.tuteehub.com/tag/cooh-409857" style="font-weight:bold;" target="_blank" title="Click to know more about COOH">COOH</a>)=("Mass"//500cm^(3)xx2)/("g.eq.mass")` <br/> Molecular wt. of `CH_(3)COOH=12xx2+1xx4+16xx2=24+04+32=60` <br/> `[CH_(3)COOH]=(3.0xx2)/60=6.0/60=0.1N` <br/> Calculation of concentration of sodium <a href="https://interviewquestions.tuteehub.com/tag/acetate-847375" style="font-weight:bold;" target="_blank" title="Click to know more about ACETATE">ACETATE</a>. `"mass"//500cm^(3)=(Nxxg" eq.mass")/2rArrN_(CH_(3)COONa)=("Mass"//500cm^(3)xx2)/("g.eq.mass")` <br/> Molecular wt. of `[CH_(3)COONa]=12xx2+1xx3+16xx2+1xx23=82` <br/> `[CH_(3)COONa]=(41xx2)/82=8.2/82=0.1N` <br/> `pH=pK_(a)+"log"(["<a href="https://interviewquestions.tuteehub.com/tag/salt-1193804" style="font-weight:bold;" target="_blank" title="Click to know more about SALT">SALT</a>"])/(["Acid"])rArrpH=-log_(10)K_(a)+"log"([CH_(3)COONa])/([CH_(3)COOH])` <br/> `pH=-log_(10)(1.75xx10^(-5))+"log"([0.1])/([0.1])=-log1.75-log10^(-5)+log1` <br/> `pH=-0.2430+5+0=4.7570`</body></html>


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