Question

You add 1 L of 100 mM HCl to 1 L of 100 mM Tris buffer (initial pH = 7.60 and a pka of 8.10). What is the approximate final pH? Select one: a. The final pH would be 12.70 b. The final pH would be 2.30 c. The final pH would be 12.08 d. The final pH would be 7.60 e. The final pH would be 7.81 f. The final pH would be 1.92 • q. The final pH would be 7.31 % h. The final pH would be 1.30 i. The final pH would be 1.42 Please check the total amount of HCl that you are adding.
please show steps. i know its not 7.60 or 7.31

Answer 1

Solution initial pn of buffer = 7.60 . pka = 8.10 using formula - 5 7.6 7.9.10 + Hoy estety ley [selt ] - - 050 [fall] = 10-0.50 - 0.3162 [acid ] C [auid] [sald] [and = ] mole fraction of 0,3162 1.0 salt = moles of salt Total males : 0:3162 0 :3162 7.0 +0.3162 ; 1.3162 = 0.2401 acid = mole of acid total miles - Jo 0.7590 bias = 0.7598 mole fraction of 4 mer S 1.3162 mole of Tris butter = = = melarity x volume 100mm x 1L. 100 xro-3 M XIL = 0.1 mol 1oo milli male Es Scanned with. CamScanner

millimale of salt in tis buffer = mole fruction of salt & mole of Tris = 0.2401 x 100 mmol = 24.01 m mol. millimale of and in tris buffer = mole fraction of acid X male of Tois = 0.7598 x 100mmol = 75.98 m modo mole of sich added = modssity x volume = 100mm x L = 100 x 103 m x 10ooml = 100 milimol added He will reach with salt fraction of Tris buffer excess he = 100 mmol - 24.01 m malo = 75.98 mmol. Total volume of solution - 1LT IL = 2L = 20 com L. concentration of ne in st ion will be due to solution = male 1. Toral volume = 75.98 nmil 2000mL = 0.037994 (mol) only and hence Tm+] = 0.037994 pH = -log[ht] pY = -log (0.037994) [ph = 1.42 ophion (i answer is wrong? Is Scanned with CamScanner Please comment it