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For each of the following solutions, calculate the initial pH and the final pH after adding...

For each of the following solutions, calculate the initial pH and the final pH after adding 0.005 mol of NaOH. 210.0 mL of a buffer solution that is 0.315 M in CH3CH2NH2 and 0.285 M in CH3CH2NH3Cl Express your answers using two decimal places separated by a comma.

Solutions

Expert Solution

Moles of base = 0.315 * 0.210 = 0.06615

Moles of salt (conjugate acid) = 0.285 * 0.210 = 0.05985

Using equation:

pOH = pKb + log(moles of salt/moles of base)

For ethylamine, pKb = 3.3

Thus,

pOH = 3.3 + log(0.05985 / 0.06615) = 3.256

Thus, initial pH = 14 - pOH = 10.744

Moles of base NaOH added = 0.005

After addition of moles of NaOH,

Moles of base in the buffer solution = 0.06615 + 0.005 = 0.07115

Moles of salt in the buffer solution = 0.05985 - 0.005 = 0.05485

Putting values in the equation:

pOH = pKb + log(moles of salt/moles of base)

pOH = 3.3 + log(0.05485 / 0.07115) = 3.187

Thus,

pH = 14 - pOH = 10.813

queen_honey_blossom answered 2 years ago
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