Question

In: Chemistry

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. 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.

Solutions

Expert Solution

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


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