Asked by Ele
How many grams of Na-benzoate (powder) do I need to prepare a buffer with pH = 4.2?
How many grams of salt (NaCl) do I need to add to this buffer to make it a saturated solution of the salt?
Thanks
How many grams of salt (NaCl) do I need to add to this buffer to make it a saturated solution of the salt?
Thanks
Answers
Answered by
DrBob222
Benzoic acid = HBz
sodium benzoate = NaBz.
Bz^- + HOH --> HBz + OH^-
Prepare an ICE chart and substitute into the following:
Kb = (Kw/Ka) = (OH^-)^2/(Bz^-)
You know Kw and Ka. Use pH to find OH^-. Solve for (Bz^-) = molar concn of sodium benzoate. How much do you wish to prepare. M = moles/L, solve for moles, then moles = grams/molar mass and solve for grams. This will not be a buffer solution, as such, but it will have a pH of 4.2. I don't know the solubility of NaCl in such a solution nor do I have any graphs that will tell me that. The addition of NaCl will not change the pH of the solution.
sodium benzoate = NaBz.
Bz^- + HOH --> HBz + OH^-
Prepare an ICE chart and substitute into the following:
Kb = (Kw/Ka) = (OH^-)^2/(Bz^-)
You know Kw and Ka. Use pH to find OH^-. Solve for (Bz^-) = molar concn of sodium benzoate. How much do you wish to prepare. M = moles/L, solve for moles, then moles = grams/molar mass and solve for grams. This will not be a buffer solution, as such, but it will have a pH of 4.2. I don't know the solubility of NaCl in such a solution nor do I have any graphs that will tell me that. The addition of NaCl will not change the pH of the solution.
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