Question

A buffer solution is prepared by placing a 25.00 mL aliquot of 0.165 M NH3 and...

A buffer solution is prepared by placing a 25.00 mL aliquot of 0.165 M NH3 and a 25.00 mL aliquot of 0.150 M NH4Cl into a 100.00 mL volumetric flask and diluting to the mark. Using activities, calculate the pH of the buffer solution. Use the Davies equation to determine activity coefficients. (pKa of NH4+ = 9.245)

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Answer #1

Volume of NH3 = 25 ml

Concentration of NH3 = 0.165 M

Volume of NH4Cl = 25 ml

Concentration of NH4Cl = 0.150ml

Final volume = 100 ml

Final concentration of NH3 = 25*0.165/100

= 0.04125 M

Final concentration of NH4Cl = 25*0.15/100

= 0.0375

Ionic strength is given by

I = \sumCizi2 = 0.0375*1^2 + 0.0375*1^2 (since NH4Cl exists as a 1:1 electrolyte with zNH4+ = 1 and zCl- = -1)

Ionic strength, I = 0.075

According to Davies equation, the activity coefficient is given by

vi -log/4 = 0.5212211 + VI - 0.3011

= 0.5*1*1(((0.075)^0.5/1+(0.075)^0.5) - 0.301*0.075)

= 0.908

Henderson-Hasselbach equation with activity coefficients is give by

[B]YB pH = pKa + log(TB 19 BH+]YBH+

where pKa is the pKa of NH4+ = 9.245

pH = 9.245 + log(0.04125/((0.0375)*0.098)) (since activity coefficient of NH3 = 1)

= 9.245 + 0.08317

= 9.328

Therefore the pH of the NH3/NH4Cl buffer including the activity coefficient = 9.328

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