You will prepare an acidic and basic buffer solutions from the following systems

Question

You will prepare an acidic and basic buffer solutions from the following systems, Basic Buffersodium hydrogen carbonate (pK 10.33)/ sodium carbonate To do so, you are disposed of the following stock solutions: 2.0 M acetic acid 2.0 M sodium acetate 1.0 M sodium hydrogen carbonate (or sodium bicarbonate) 1.0 M sodium carbonate Each buffer needs to be prepared by mixing 50 mL of a dilute weak acid solution with another 50 mL of a dilute solution of its conjugate species, so that the total volume is 100 mL and the individual concentrations of the weak acid and weak base conjugates in the buffer is 0.10 M B.1 Derive in each case, the concentration that the weak acid and weak base species should be B.2 Based on your answer from part B.1, calculate the volume of stock solution that you need to dilute B.3 Based on your answer from part B.2, dilute appropriate aliquots of the weak acid and weak base prepared at prior to mixing. in order to make 100 mL of the individual weak acid and weak base solutions. stock solutions, separately in 100 mL volumetric flasks.

Explanation / Answer

According to Henderson-Hasselbach formula

pH = pKa + log [(acetate)/(acetic acid)]

For given 2.0 M solutions of sodium acetate and acetic acid with pKa = 4.74

Substituting the above values into Henderson-Hasselbach formula one will get the pH of solution as follows:

pH = 4.74 + log [(2.0)/(2.0)]

pH = 4.74 + log [1]

pH = 4.74 + 0

pH = 4.74

As can be seen from above at equal concentrations of acid and its conjugate base the buffer pH equals to that of pKa

Accordingly for the buffer solution prepared from 0.1 M solutions of weak acid and weak base the pH will be equal to pKa of that weak acid.

Answer for B1:

So for a final concentration of 0.1 M of 100 mL solution the individual initial concentration of stock solutions of acetic acid and sodium acetate assumed to be the given concentration of 2 M.

Similarly in case of sodium hydrogen carbonate and sodium carbonate the individual initial concentration is assumed as the given initial concentration, which is 1M.

Answer for B2:

So for a final concentration of 0.1 M and a stock solution of 100 mL solution the initial individual volume of acetic acid and sodium acetate can be calculated as follows:

concentration of final solution, M2 = 0.1 M

final volume of buffer solution, V2 = 100 mL

Assumed initial concentration of each stock solution is, M1 = 2.0 M

So the initial volume, V1 to be used = M2 x V2 / M1 = 0.1 x 100 / 2 = 5 mL

So for a concentration of final 0.1 M of 100 mL solution the individual volume of stock solutions of acetic acid and sodium acetate should be of 5 mL.

In case of sodium hydrogen carbonate and sodium carbonate buffer the individual volume of stock solution of each of sodium hydrogen carbonate and sodium carbonate to be used can be calculated as follows.

concentration of final solution, M2 = 0.1 M

final volume of buffer solution, V2 = 100 mL

Assumed initial concentration of each solution is, M1 = 1.0 M

So the initial volume, V1 to be used = M2 x V2 / M1 = 0.1 x 100 / 1 = 10 mL

So for a concentration of final 0.1 M of 100 mL solution of sodium hydrogen carbonate and sodium carbonate buffer the individual volume of stock solution of each of sodium hydrogen carbonate and sodium carbonate to be used is 10 mL.

Answer for B3:

So for a concentration of final 0.1 M of 100 mL solution the individual volume of acetic acid and sodium acetate should be of 5 mL. That is a total volume of 10 mL of mixed solutions which is further diluted to 100 mL by adding 90 mL of deionized water to obtain the desired 0.1M buffer solution of 100 mL

In case of a final concentration of 0.1 M of 100 mL solution of sodium hydrogen carbonate and sodium carbonate buffer the individual volume of stock solution of each of sodium hydrogen carbonate and sodium carbonate to be used is 10 mL. That is a total volume of 20 mL of mixed solutions which is further diluted to 100 mL by adding 80 mL of deionized water to obtain the desired 0.1M buffer solution of 100 mL.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.