Question

14 of Soin. when 1.000 of O.SDOM acetic acid ? NaOH is dissolved in 0.00 TL pka - 4.74 of acetic acid NaOH + CH3COOH Igx Imol CH₃COONa+H2O - 39.997 = 0.025 mol 0.SOOM * 0.1L = 0.05 mol acest acid NaoH 0.05-0.025 = 0.025 mol excess NaOH. 0.025 mol 0.25M NaOH = COH) pOH = -log (0.25m) = 0.602 pH = 14- 0.602 = 13.398

wondering if this is the correct way to solve this problem

Answer 1

NaOH + CH3COOH (base) (acid) CH3 Coowa + H20 (salt) (water) The morar mass of waoH = (23+16+1) = 40 g/mon 1.00, waoH = yo gimo = 0.025 moi waoh 3 100 ML = 100x10 L of 0.500 M acetic acid contains (0.500 x 100x103) = 0·05 mo of acetic acid. NaOH + CH3 C CH3Coona + H20 0.025 0.05 moo o Initial: equibrium: 0.025mor 0.025mon 0.025 mos le 0.025 mor of waost and 0.025 mos of acetic acid reacts to produce 0.025 mol sodium acetate as a salt. hear. Limiting reagent is waoH. Rema ning CH3 = (0.05-0.025) mo = 0.025 mor so this is the acetic buter socution as aless 0.025 mor acetic acid remain.

Total vaume of solution - 100 ml = 1008103 L Moanity of acetic acid = 0.025 mo 100x10 3 L = 0.25 mol Maarity of sodium acetate = 0.025 mor 1008103 L = 0.25 morlL According to Henderson equation, the pa of acetic buster solution is, (Csalt- concentration of salt pH = Pka + og Csalt that cacid pH = 4.74 + rog 0125 V 0.25 (pka = 4.74 (Pka = 4.74 ot acctic acid) pH = 4.74 :( log1=0) pH of the solution is 4.74