Question

How can you prepare 100.0 mL of acetic buffer solution (0.100 M) with pH = 5.10 by using a solution of acetic acid (25% w/w, density = 1.05 g/cm3 ) and pure sodium acetate. (pKa = 4.76; M. W. = 60.05 g/mol for HAc and 82.03 for NaAc).

Answer 1

A buffer solution is a mixture of weak acid and its salt with strong base or vice versa.

Here weak acid is HAc and its salt with strong base is NaAc

pH for buffer solutions is given by henderson hasselbalch equation

pH = pKa + log(salt/acid)

Where pKa is already given to us as 4.76 and salt/acid is the ratio of concentration of salt and acid

We have to prepare a solution where pH = 5.1 so,

5.1 = 4.76 + log(salt/acid)

We are also given that concentration of salt [NaAc] + acid [HAc] = 0.1 M

so, log([NaAc]/[HAc]) = 5.1-4.76 = 0.34

so, [NaAc]/[HAc] = 10^0.34 = 2.1877

For 25% w/w solution of HAc we have 25g of HAc dissolved in 100 g of solution.

As density d1 = 1.05g/cm3 = 1.05g/ml so, 100 g solution means volume of 100g solution = mass/density = 100 g/ 1.05 g/ml = 95.238 ml = 95.238/1000 L = 0.095238 L

Also, no. of moles of HAc in 25g = weight/molecular weight = 25g/60.05g/mol = 0.416 mol

so, molarity of HAc = moles/volume in L = 0.416/0.095238 M = 4.368 M

In 100 ml from this solution we will have 4.368*100 millimoles = 436.8 millimoles of HAc in the solution

As we have [NaAc]/[HAc] = 2.1877 we need 2.1877*[HAc] moles or 2.1877*436.8 millimoles of NaAc = 955.587 millimoles = 955.587/1000 moles = 0.955587 moles

Each mole of NaAc has weight of 82.03 g so mass of 0.955587 moles of NaAc will be 0.955587 * 82.03 g = 78.3868 g (mole = weight/molecular weight)

So, we can prepare the required solution by taking 100 ml of acetic acid solution and then adding 78.3868 g of pure NaAc in it.