Question 6. 1M CH3COOH solution. The dissociation constant of acetic acid is 1.8 x 10 -5 . Answer: pH = – log[H + ] For weak acids,
Question 7. Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. Ka for acetic acid is 1.8 x 10 -5
Given that Ka = 1.8 x 10 -5 pKa = – log (1.8 x 10 -5 ) = 5 – log 1.8 = 5 – 0.26 = cuatro.74 pH = 4.74 + log \(\frac < 0.20>< 0.18>\) = 4.74 + log \(\frac < 10>< 9>\) = 4.74 + log 10 – log 9 = 4.74 + 1 – 0.95 = 5.74 – 0.95 = 4.79
Concern 8. What is the pH off an enthusiastic aqueous provider obtained by combo 6 gram from acetic acid and you may 8.dos gram off sodium acetate and you can putting some regularity comparable to 500 ml. (Given: K to possess acetic acidic is 8 x 10) Answer: Predicated on Henderson – Hessalbalch picture,