Question

What is the pH of a buffer that is 0.379 M HF and 0.374 M LiF? The Ka for HF is 3.5 x 10-4. If the ΔH = 109 kJ/mol and ΔS = 100 J/K mol, at what temperature does this reaction become spontaneous?

Answer 1

By using Henderson -- Hasselbalch Equation PH = Pka + log TAB Given – HF = 0.379 M. Lif = 0.374 M 7.5 X 0-4 Ka = 00374 0.379 Hf & Attf FFT Pn = Pka + log The pH = -log (3.5 X10~4) + log 0. PK = 3.46 - 0.0050 PH = 3.45 Please Note that By wing log -log (3-5 X 1079) = 3.46 -10 12.64-47 log 0:339 =-0.0058 table e By using 19 = 4H-TAS Ak = 109 kJ/mol. As = 1.00 J/K = 0.1 K1/k if AG=0 then the System is at the limit of reaction Spontaneity

ANT 0 = 109 kJ/mol - TX 0-1 KJ/K TX 0-1 = 109 109 T= 1090 K, TE 87% Thank hobe this will ill help you
therefore the reaction becomes spontaneous when T = 1090K ( 817ºC)