Question

# 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 0.0100 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.

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).

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).

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

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