Use the Hendersen-Hasselbalch equation to calculate the amount of Na+CH3COO- and

Question

Use the Hendersen-Hasselbalch equation to calculate the amount of Na+CH3COO- and CH3COOH needed to make a 100 mL solution of 50 mM acetate/acetic acid with a pH of 4.0. For this part 0.2 M acetic acid will be the stock solution. Please show work. [MW of acetic acid = 60.05 g/mol, MW of sodium acetate = 82.04 g/mol, density of acetic acid = 1.049 g/mL, pKa of acetic acid = 4.757]

My question is, why is the molarity of the stock solution not used in the calculation?

Here is how I did the problem:

[A-] + [HA] = 50 mM  0.05 M
[A-] = 0.05 M - [HA]   you must plug this value into the Hendersen-Hasselbalch equation

ph = pka + log ([A-] / [HA])
4.0 - 4.757 = log ((0.05 M - [HA]) / [HA])
1.175[HA] = 0.05 - [HA]
[HA] = 0.0425 M  to convert to moles you need to multiply by how much you are using which in this case is 100 ml or 0.1 L

0.0425 moles/L (molarity) x 0.1 L = 0.00425 moles of Acetic Acid x 60.05 g/mol x 1mL/1.049 (density) = 2.433 mL of Acetic Acid

Now you can take the derived molarity of Acetic Acid to get molarity of Acetate using the established relationship equation from before.

[A-] + [HA] = 0.05 M
[A-] = 0.05M - 0.0425 M
= 0.0075 M x 0.1L = 0.00075 moles of Acetate salt
0.00075 moles x 82.04 g/mol of sodium acetate = 0.0615 g

Explanation / Answer

The molarity of the stock solution is given because we are supposed to dilute the stock solution to arrive at the buffer solution. I will do the problem in a slightly different manner.

The total molarity of the solution is 50 mM = 0.05 M

Therefore, [HAc] + [Ac-] = 0.05 M ….(1)

Also, from the Henderson-Hasslebach equation,

pH = pKa + log10[Ac-]/[HAc]

==> 4.0 = 4.757 + log10[Ac-]/[HAc]

==> -0.757 = log10[Ac-]/[HAc]

==> [Ac-]/[HAc] = 10-0.757 =0.1749

==> [Ac-] = 0.1749[HAc] ….(2)

Equations (1) and (2) are simultaneous equations in two unknowns. We solve them.

From (2) in (1),

[HAc] + 0.1749[HAc] = 0.05 M

==> 1.1749[HAc] = 0.05 M

==> [HAc] = 0.0425 M

[Ac-] = 0.05 M – 0.0425 M = 7.5*10-3 M

So far everything works out fine. However, we must realize that it is difficult to weigh out acetic acid (as it is a liquid) and dilution of a stock solution is a better and easier approach. Infact, in this case, we are supposed to do a dilution only; hence the stock solution is given.

Moles of acetic acid in 100 mL = 0.1 L of the buffer solution is (0.1 L)*(0.0425 mol/L) = 4.25*10-3 mol.

Dividing the required number of mole(s) by the molarity of the stock solution will give the volume of acetic acid required to prepare the buffer.

Volume of acetic acid required = 4.25*10-3 mol/(0.2 mol/L) = 0.02125 L = 21.25 mL

The volume of stock acetic acid solution that needs to be taken = 21.25 mL (ans)

Next, we calculate the amount of sodium acetate taken. Sodium acetate is a solid and can be easily weighed.

The given volume of buffer contains (0.1 L)*(7.5*10-3 mol/L) = 7.5*10-4 mole sodium acetate.

Weight of sodium acetate that needs to be weighed out = (7.5*10-4 mol)*(82.04 gm/mol) = 0.06153 gm

The weight of sodium acetate that needs to be weighed out = 82.04 gm (ans)

The density and molar mass of acetic acid are redundant information; at best we can use the density to calculate the weight of acetic acid that has to be taken. This can be done as

Weight of acetic acid taken = 21.25 mL* 1.049 gm/mL = 22.29 gm.

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