Describe the preparation of 1.5 liters of a 150 mM acetate buffer pH 4.00. The m

Question

Describe the preparation of 1.5 liters of a 150 mM acetate buffer pH 4.00. The materials you are allowed

to use to make this solution are: 2.0 M sodium acetate stock, 6 M HCl stock and the necessary volume

measuring devices and pH meter. Show your calculations and explain in excruciating detail how you would

make up the buffer step by step, including any dilutions of stock solutions you might make to make volume

measurements more accurate. Be sure to describe the glassware and measuring devices you would use to

make this buffer as accurately as possible.

Explanation / Answer

The given reagents are 2.0 M sodium acetate (NaC2H3O2) and 6 M HCl.

NaC2H3O2 is the conjugate base of the weak acid, acetic acid (HC2H3O2). The buffer consists of the weak acid (HC2H3O2) and its sodium salt (NaC2H3O2); however, we are provided with NaC2H3O2 only and HCl. HC2H3O2 is prepared insitu by the reaction between NaC2H3O2 and HCl as below.

NaC2H3O2 (aq) + HCl (aq) --------> NaCl (aq) + HC2H3O2 (aq)

As per the stoichiometric equation above,

1 mole NaC2H3O2 = 1 mole HCl = 1 mole HC2H3O2 …...(1)

The buffer system is denoted by the reaction below.

HC2H3O2 (aq) --------> H+ (aq) + C2H3O2- (aq); pKa = 4.76 at 25ºC.

C2H3O2- is present as the sodium salt, i.e, NaC2H3O2.

Determine the ratio of the concentrations of HC2H3O2 and NaC2H3O2 in the buffer using the Henderson-Hasslebach equation as below.

pH = pKa + log [NaC2H3O2]/[HC2H3O2]

====> 4.00 = 4.76 + log [NaC2H3O2]/[HC2H3O2]

====> -0.76 = log [NaC2H3O2]/[HC2H3O2]

====> [NaC2H3O2]/[HC2H3O2] = antilog (-0.76)

====> [NaC2H3O2]/[HC2H3O2] = 0.1738

====> [NaC2H3O2] = 0.1738*[HC2H3O2] ……..(2)

The total acetate concentration in the buffer is given as 150 mM; therefore,

[NaC2H3O2] + [HC2H3O2] = 150 mM

====> 0.1738*[HC2H3O2] + [HC2H3O2] = 150 mM

====> 1.1738*[HC2H3O2] = (150 mM)/(1.1738)

====> [HC2H3O2] = 127.7901 mM 127.8 mM.

[NaC2H3O2] = 0.1738*[HC2H3O2] = 0.1738*(127.8 mM)

=22.21164 mM 22.2 mM.

Since, we have 1.5 L of the buffer, we can determine the moles of HC2H3O2 and NaC2H3O2 in the buffer. Therefore,

Moles of HC2H3O2 in the buffer = (1.5 L)*(127.8 mM) = (1.5 L)*(127.8 mM)*(1 M)/(1000 mM)

= 0.1917 mole.

Moles of NaC2H3O2 in the buffer = (1.5 L)*(22.2 mM) = (1.5 L)*(22.2 mM)*(1 M)/(1000 mM)

= 0.0333 mole.

Note that HC2H3O2 wasn’t present originally and was prepared by partial neutralization of NaC2H3O2 with HCl. Therefore,

Moles of NaC2H3O2 neutralized = moles of HCl used = moles of HC2H3O2 formed = 0.1917 mole (as per stoichiometric equation 1 above).

Therefore, moles of NaC2H3O2 originally present = (moles of NaC2H3O2 present in the buffer) + (moles of NaC2H3O2 neutralized) = (0.0333 mole) + (0.1917 mole) = 0.2250 mole.

Now, the stock solution of NaC2H3O2 had concentration 2.0 M; therefore, volume of the stock solution required to prepare the buffer = (total number of moles of NaC2H3O2 originally present)/(molarity of stock solution)

= (0.2250 mole)/(2.0 M)

= 0.1125 L (1 M = 1 mol.L-1)

= (0.1125 L)*(1000 mL)/(1 L) = 112.50 mL.

Moles of HCl added = moles of HC2H3O2 formed = 0.1917 mole; therefore, the volume of 6 M HCl required = (moles of HCl required)/(molarity of stock HCl)

= (0.1917 mole)/(6 M)

= 0.03195 L

= (0.03195 L)*(1000 mL)/(1 L)

= 31.95 mL.

Therefore, the buffer is prepared by adding 112.50 mL of stock 2.0 M sodium acetate and 31.95 mL of 6 M HCl in the 1.5 L volumetric flask and diluting upto the mark with deionized water.

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