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For 510.0 mL of a buffer solution that is 0.170 M in CH 3 CH 2...

For 510.0 mL of a buffer solution that is 0.170 M in CH 3 CH 2 NH 2 and 0.150 M in CH 3 CH 2 NH 3 Cl , calculate the initial pH and the final pH after adding 0.010 mol of HCl . Express your answers using two decimal places separated by a comma.

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Expert Solution

pKb of ethylamine = 3.19

According to the Hinderson Hasselbalch equation :

pOH = pKb + log [CH3CH2NH3Cl/CH3CH2NH2]

       = 3.19 + log [0.15/0.17]

         = 3.14

Initial pH of the solution = 14-pOH = 14-3.19 = 10.81

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when HCl is added, it will react with ethyl amine to convert it to ethyl ammonium chloride

moles of acid added = moles of ethyl ammonium chloride formed. This amount has to be added to ethyl ammonium chloride intially present. same has to be substracted from ethyl amine intitally present

moles of ethyl amine initially present = 0.17 M * 0.510 L = 0.0867 moles

Moles of ethylammonium chloride present = 0.15 M * 0.510 L = 0.0765 moles

Moles of HCl added = 0.01 moles

Moles of ethylamine present after adding HCl = 0.0867 moles - 0.01 = 0.0767 moles

Moles of ethylammoniumchloride present after adding HCl = 0.0765 moles + 0.01 = 0.0865

pOH = pKb + log [CH3CH2NH3Cl/CH3CH2NH2]

       = 3.19 + log [0.0865/0.0767] = 3.24

pH of the solution = 14-pOH = 14-3.24 = 10.76

queen_honey_blossom answered 2 years ago
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