If you prepared a 0.15 M CH3COOH - 0.15 M CH3CO2Na buffer, the resulting buffer capacity would be expected to be the same as the buffer you prepared in Part II of this lab.
Explanation:
In Part II of this lab, moles of CH3COOH = 0.5 mol/L * (10/1000) L = 0.005 mol
And moles of CH3COONa = 0.5 mol/L * (10/1000) L = 0.005 mol
Now, pH = pKa + Log([CH3COONa]/[CH3COOH])
i.e. pH = 4.74 + Log(0.005/0.005)
i.e. pH = 4.74
The given buffer does also have the same pH. Hence, they have the same buffer capacity.
2. If you prepared a 0.15 M CH.COH - 0.15 M CHCO Na buffer, would you expect the resulting buffer capacity t...
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