1) a buffer is prepared with 0.15 M CH3COO- and 0.57 M CH3COOH. What is the PH?

Question

1) a buffer is prepared with 0.15 M CH3COO- and 0.57 M CH3COOH. What is the PH? The ka for CH3COOH is 1.75x 10^-5

2) a buffer is prepared with 0.27 M NH4+ and 0.17 M NH3. What is the PH? The ka for NH4+ is 5.6x10^-10? 1) a buffer is prepared with 0.15 M CH3COO- and 0.57 M CH3COOH. What is the PH? The ka for CH3COOH is 1.75x 10^-5

2) a buffer is prepared with 0.27 M NH4+ and 0.17 M NH3. What is the PH? The ka for NH4+ is 5.6x10^-10?

2) a buffer is prepared with 0.27 M NH4+ and 0.17 M NH3. What is the PH? The ka for NH4+ is 5.6x10^-10?

Explanation / Answer

1

For the ethanoic acid there is an equilibrium between the unionized acid and its ions:

CH?COOH (aq) ?CH?COO? (aq) + H? (aq)

so Ka = [CH?COO?] * [H?] / [CH?COOH]



Ethanoate ions are added so to keep the equilibrium the reaction goes to the left side ... [using Le Chatelier's Principle] ... and the final [CH?COO?] is mostly due to the added ethanoate ions ... but the equilibrium is maintained and the equation for Ka is the same




now have to get the concentrations:

Ethanoic acid is a weak acid ... so the ethanoate ions from the ethanoic acid is negligible compared to those added directly ... so assume that [CH?COO?] = concentration of the added ethanate ...

so [CH?COO?] = 0.15M


B/c CH?COOH is a weak acid very little ionization will occur ... so assume that the concentration of the acid at equilibrium is the same as the concentration of the acid used ...

so CH?COOH = 0.57M

Ka = 1.75 x 10??

so [CH?COO?] * [H?] / [CH?COOH] = 1.75 x 10??

subsing:

0.15 * [H?] / 0.57 = 1.75 x 10??

[H?] = 6.65 x 10??

pH = -log [H?]

= -log (6.65 x 10??)

= 4.18

so the pH is 4.2 (to 2 sig figs)

2

The Henderson-Hasselbalch equation is:
pH = pKa + log(A/AH)
In this case A = NH3and AH = NH4+. The pKb of NH3 is 4.75. Substituting,
pH = (14-4.75) + log(0.27 / 0.17) = 9.26

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