Question

Calculate the pH of a buffer made from mixing 7.6 mL of 0.259 M NaC2H3O2 and 12.1 mL of 0.338 M HC2H3O2. The Ka of HC2H3O2 is 1.8 x 10-5.

Answer 1

7.6 mL of 0.259 (M) NaC₂H₂O2 is mixed with 12:1 mL of 0.338(M) HC2H30g acid). The total volume of the solution becomes (7.6+12:10mi = 19.7 ml Note - If we change the volume of s, concentration solution to the new volume of V₂ by any means. (like addition of more solvent to the solution). Then, the concentration of the solution also changes from si to Sq due to the change in solvent volume. They are related as follows: . us, = V₂S, : Sq = . The concentration of Na CzHz Oq be in 19.7ml solution becomes = [NaC₂H₂ O₂] = 7.64Lx 0,258(M) 1907 hat .:[NaC, H302] = 0.0999(M); 2 similarly the concentration of Hetzlz in igitmi solution becomes = [+Q2h302] = 12.1 GLX 0.338(M). 19.7 mx. ... [HC, H3 De] => 0.2076(M) Henderson - Hasselbalch equation for acidiebuffer is pH = pkat log Isalt] o Facidy bei [salt

...pH = pka + log [NaC H₂02 ] | Kaz.18x105. [HC2 H₂ Oz ] Pkaa-log (1.8*205) Pka z 4.7447 :. Ph - 4.744 + Lºa 69) 10. 2076) ; 4.7447+ 60.3177) = 4.7447 -0.3177 = 4.427 . . i pr of the buffer made from mixing of 7.6 m2 of 0.259 (M) NaCa H 3 Oq and 12.1 ml of 0.338(M) He H 3 O₂ is 4:427.