Question

1. Calculate the pH of a buffer solution prepared by mixing 0.250 L of 0.150 M acetic acid with 0.200 L of 0.250 M sodium acetate. Ka acetic acid = 1.8 x 10-5 Calculate the pH of this solution of after the addition of 0.003 L of 0.200 M HCL 2. Consider a buffer solution prepared by mixing 0.250 L of 0.150 M acetic acid with 0.200 L of 0.250 M sodium acetate. Ka acetic acid = 1.8 x 10-5 Calculate the pH of this solution of after the addition of 0.003 L of 0.200 M NaOH. 3. Calculate the pH of a solution prepared by the addition of 0.003 L of 0.200 M HCl to pure water.

Answer 1

We are mixing 0.25L of 0.150M acetic acid = 0.254 x 0.15 mol = 0.0375 moles of Acid

VU V to 0.21 of 0.25M Sodium Acetate (Nao Ac) = 0.IM +0.24x. 0.25 mol Ka (Acid) = 1.8xlos pkä -log (1.8xos) = 0.05 mol of Nadac (salt) +0.2L I kā 4.745 After dilution total volume = 0.25 (Mixing) - 0.45L pH = pka + log [salt] Naotc id] ACOH Conc. of Acid in new volume = 0.0375 mol 10.45 L PH = 4.745 + logollllM' 10.0833m) [PH = 4.87 = 0.08333M Cone of salt in new volume = 0.05 mol Re0832 0.45

moles = 0.0375 +68/6" 0.0381 moles. New Conc. of ACOH = 0.0381 moles 0.453L Now to this solution i.e, 0.45L of solection has 4 0.0633M Acetic acid (ACH) 4 0.IM Sodium Acetate (NaoAc) PH=4.87 To this solution, we are adding 0.003L of 0.20 M HCl 0.0034 x 0.20 mol = 6*10-4 mol of HCl Reaction taking place! 0.084106M Now, PH can be calculated, pH = pka + log [salt] (Acid = 4.745 + log 6. 10905) 16.084106 IPH=4.858 CH₃Coona + HCl CH₂ Cool + Nach &. 1 mol HCl react with 1 mol of Nadal This pH of solution cefter HCl addition was found to be 4.858 : 6X104 mol " " 6xlo "mol of Massac and produce 6 Xlot mol of ACOH (NaOAC) moles = Original moles - 6xlo "mol. = 0.05- 6x104 = 0.0494 moles. final volume becomes= 0.451+0.0032 = 0.453 L New concentration = of Naote 0.6494 mol 0.453 L = 0. 10905 M Similarly (AcoHandles - Original + 6x10" moles