Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH.

Part A

For 300.0 mL of pure water, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH.

Express your answers using two decimal places separated by a comma.

Part B

For 300.0 mL of a buffer solution that is 0.210 M in HCHO2 and 0.280 M in KCHO2, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH(Ka=1.8⋅10−4).

Express your answers using two decimal places separated by a comma.

Part C

For 300.0 mL of a buffer solution that is 0.2551 M in CH3CH2NH2 and 0.2251 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH(Kb=5.6⋅10−4).

Express your answers using two decimal places separated by a comma.

Answer 1

Part A)

Initial pH = 7

Moles of NaOH added = 0.01

Volume of solution = 300 mL = 0.3 L

=> [NaOH] = 0.01 / 0.3 = 0.0333 M (For all the parts)

NaOH is a strong base. Hence, it dissociates completely in aquous solution

[OH-] from NaOH = 0.0333 M

pOH = - log [OH-] = - log (0.0333) = 1.477

pH = 14 - pOH = 12.52

Part B)

Ka = 1.8 x 10^-4

=> pKa = 3.745

Initial pH = pKa + log ([KCHO2] / [HCHO2])

=> pH = 3.745 + log (0.28 / 0.21) = 3.87

Final pH

[NaOH] = 0.0333 M

Final pH = pKa + log ([KCHO2] + [NaOH] / [HCHO2] - [NaOH])

pH = 3.745 + log (0.28 + 0.0333 / 0.21 - 0.0333) = 3.99

Part C)

pKb = 3.25

Initially

pOH = pKb + log ([CH3CH2NH3Cl] / [CH3CH2NH2])

=> pOH = 3.25 + log (0.2251 / 0.2551) = 3.20

pH = 14 - pOH = 10.80,

Finally

pOH = pKb + log ([CH3CH2NH3Cl] - [NaOH] / [CH3CH2NH2] + [NaOH])

=> pOH = 3.25 + log (0.2251 - 0.0333 / 0.2551 + 0.0333) = 3.07

pH = 14 - pOH = 10.93