A buffer solution of pH = 4.7 is prepared from CH_(3)COONa and CH_(3)COOH. Dissociation constant of acetic acid is 1.75xx10^(-5). Calculate the mole proportion of sodium acetate and acetic acid.

A buffer solution of pH = 4.7 is prepared from CH_(3)COONa and CH_(3)C

1.	A buffer solution of pH = 4.7 is prepared from CH_(3)COONa and CH_(3)COOH. Dissociation constant of acetic acid is 1.75xx10^(-5). Calculate the mole proportion of sodium acetate and acetic acid.
Answer» <html><body><p></p>Solution :`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_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)COONa])/([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>])` <br/> `4.7=-log_(10)(1.75xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>))+"log"([CH_(3)COONa])/([CH_(3)COOH])` <br/> `4.7=5-log1.75+"log"([CH_(3)COONa])/([CH_(3)COOH])=5-0.2430+"log"([CH_(3)COONa])/([CH_(3)COOH])` <br/> `"log"([CH_(3)COONa])/([CH_(3)COOH])="anti"log(-0.0557)="anti"log(-0.057+1-1)` <br/> `([CH_(3)COONa])/([CH_(3)COOH])=0.8770`.</body></html>