1) What combination of a molecular base and a salt of the conjugate acid of the
ID: 833714 • Letter: 1
Question
1) What combination of a molecular base and a salt of the conjugate acid of the molecular base should you select to prepare a buffer solution that would maintain a pH of approximately 9.50? Please Explain.
1a) What ratio of concentrations of the molecular base you selected and its conjugate acid would be necessary to achiece this pH?
1b) Describe how you would make 1.00 L of this buffer solution. Assume that 1.0 M solution of the molecular base is available and that the chloride salt of the conjugate acid is available as a solid.
Explanation / Answer
Ammonia! It has a Ka of about 5.56e-10. So it's pKa = -log(Ka) = 9.25 so it is very close to the pH you want. Ammonia's formula is NH3 so it fits the molecular requirement. It's conjugate acid is NH4+. One of it's salts is very common: NH4Cl.
Now to figure out how much of each we need let us use the henderson hasselbach equation:
pH = pKa + log([base]/[acid])
9.5 = 9.25 + log([base]/[acid])
log([base]/[acid]) = 0.25
[base]/[acid] = 10^0.25 = 1.778
So the ratio of base/acid = 1.771/1.
Now to describe how to make the buffer:
First we have to figure out what concentration of the base we need in the final solution. Let's just say we want the final concentration of the base to be 0.1M.
Then we use the equation M1V1 = M2V2; 1 * V1 = 0.1 * 1; V1 = 0.1
So we add 0.1 liters of the 1M base solution. Then to figure out how much of the acid we need:
base/acid = 0.566 (from above); so acid = base/0.566 = 0.1/1.778 = 0.0562M
Molarity = moles/Liters so moles = Molarity * volume which in this case = 0.0562 * 1
So we need 0.0562 moles of the acid salt.
How heavy is NH4Cl? Its molecular weight is 53.5 g/mol
0.0562mol * 53.5 g/mol = 3.008g.
So the saying is never add water to acid so the best way to make the buffer now knowing everything we need is to get about 0.5L of water then add 0.1L of the 1M NH3 solution, then add 3.008g of NH4CL, then add water until the final volume is 1L and then you have your solution at pH 9.5.
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