Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.005 mol of NaOH. Separate your answers using a comma.

Part A: For 260.0 mL of pure water, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.

Part B: For 260.0 mL of a buffer solution that is 0.245 M in HCHO2 and 0.275 M in KCHO2, calculate the initial pH and the final pHafter adding 0.005 mol of NaOH.

Part C: For 260.0 mL of a buffer solution that is 0.260 M in CH3CH2NH2 and 0.230 M in CH3CH2NH3Cl, calculate the initial pHand the final pH after adding 0.005 mol of NaOH.

Answer 1

Part A)

Initial pH = 7

Moles of NaOH added = 0.005

Volume of solution = 260 mL = 0.26 L

=> [NaOH] = 0.005 / 0.26= 0.0192 M (For all the parts)

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

[OH-] from NaOH = 0.0192 M

pOH = - log [OH-] = - log (0.0192) = 1.716

pH = 14 - poh = 12.28

Part B)

Ka = 1.8 x 10^-4

=> pKa = 3.745

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

=> pH = 3.745 + log (0.275 / 0.245) = 3.79

Final pH

[NaOH] = 0.0192 M

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

pH = 3.745 + log (0.275 + 0.0192 / 0.245- 0.0192) = 3.86

Part C)

(Kb=5.6⋅10−4).

pKb = 3.25

Initially

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

=> pOH = 3.25 + log (0.230 / 0.260) = approximately 3.20

pH = 14 - pOH = 10.80,

Finally

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

=> pOH = 3.25 + log (0.230 - 0.0192 / 0.260+ 0.0192) = 3.13

pH = 14 - pOH = 10.87