Homework Answers
a)
no of mol of CH3COONa taken in solution 1 = 0.1*0.25 = 0.025 mol
No of mol of HCl added = CH3COOH formed = 0.05*0.25 = 0.0125 mol
finally no of mol of CH3COONa present in solution 1 = 0.025 - 0.0125 = 0.0125 mol
pH of acidic buffer = pka + log(ch3coona+strong base/ch3cooh-strong base)
pka of ch3cooh = -logKa = -log(1.78*10^-5) = 4.75
No of mole of acetic acid present in buffer = 0.0125 mole
No of mole of sodium acetate = 0.0125 mole
No of mole of strong base added = 4.5/1000 = 0.0045 mole
pH = 4.75 + log((0.0125+0.0045)/(0.0125-0.00455))
pH = 5.08
b)
no of mol of CH3COOH taken in solution 1 = 0.1*0.25 = 0.025 mol
No of mol of NaOH added = CH3COONa formed = 0.05*0.25 = 0.0125 mol
finally no of mol of CH3COOH present in solution 1 = 0.025 - 0.0125 = 0.0125 mol
pH of acidic buffer = pka + log(ch3coona+strong base/ch3cooh-strong base)
pka of ch3cooh = -logKa = -log(1.78*10^-5) = 4.75
No of mole of acetic acid present in buffer = 0.0125 mole
No of mole of CH3COONa = 0.0125 mole
No of mole of strong base added = 4.5/1000 = 0.0045 mole
pH = 4.75 + log((0.0125+0.0045)/(0.0125-0.00455))
pH = 5.08
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