Buffer -1 will have strong action and better PH resisting capacity. If constituents of buffer are in concentrated form it will have strong buffering action.it can be shown through calculation also using Henderson equation
PH = Pka + log [ CH3COO- ] / [
CH3COOH]
Let we have taken 1 litre of both the buffer solution.
FOR BUFFER SOLUTION-1
initial PH = Pka = 4.74
when 0.06 mol of strong base is added it will neutralise 0.06 mol CH3COOH and form extra 0.06 mol acetate ion according to reaction.
CH3COOH + OH- CH3COO- + H2O
so moles of CH3COOH = 0.50 - 0.06 = 0.44 mol
moles of acetate ion = 0.50 + 0.06 = 0.56 mol
PH = 4.74 + log ( 0.56 / 0.44 ) = 4.74 + 0.1 = 0.75
change in PH = 0.1
For BUFFER -2
initial PH = 4.74
when 0.06 mol of base is added it will completely neutralise 0.05 mol acetic acid and form extra 0.05 mol acetate ion . So final solution will have 0.1 mol acetate ion and remaining 0.01 mol strong base.which is definitely not a buffer solution.
Acetate buffer-1 and buffer-2 were developed by your colleague. Acetate buffer 1 is composed of 0.50...
1) Calculate the pH of the 1L buffer composed of 500 mL of 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of HCl is added (Ka HC2H3O2 = 1.75 x 10-5). Report your answer to the hundredths place. 2) Calculate the pH of the 1L buffer composed of 500 mL 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of NaOH is added (Ka HC2H3O2 = 1.75...
Consider a buffer composed of the weak acid acetic acid (CH3COOH) and the conjugate base sodium acetate (NaCH3COO). Which pair of concentrations results in the most effective buffer (i.e. has the highest buffer capacity)? A. 0.10 M CH3COOH; 0.10 M NaCH3COO B. 0.90 M CH3COOH; 0.10 M NaCH3COO C. 0.10 M CH3COOH; 0.90 M NaCH3COO D. 0.50 M CH3COOH; 0.50 M NaCH3COO
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