The final volume of buffer solution must be 100.00 mL and the final concentratio
ID: 1026653 • Letter: T
Question
The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M. Based on this information, what mass of solid conjugate base should the student weigh out to make the buffer solution with a pH-2.5? Based on this information, what volume of acid should the student measure to make the 0.100 M buffer solution? Number Number 7.73 239.9 mL Incorrect. Rearrange the equation below to determine the necessary concentration of conjugate base, or [A1 Then, use the molar mass of the solid salt to calculate the mass that should be weighed out.Explanation / Answer
The Ka values of the acids (1st Ka values in case of polyprotic acids) are given. Determine pKa values as pKa = -log Ka
Weak Acid
Ka
pKa
Phosphoric acid, 1.00 M
7.52*10-3
2.12
Formic acid, 2.00 M
1.77*10-4
3.75
Sodium disulfate monohydrate, 3.00 M
1.20*10-2
1.92
Acetic acid, 5.00 M
1.75*10-5
4.76
The weak acid is so chosen that the pKa of the weak acid lies within unit of the pH. Since (pH – 1) < pKa < (pH +1) for phosphoric acid, hence, phosphoric acid is the best choice. The buffer is prepared by adding phosphoric acid and its conjugate base, sodium dihydrogen phosphate monohydrate, NaH2PO4.H2O.
The supplied phosphoric acid is 1.00 M; the buffer is 100 mL and the acid concentration in the buffer is 0.100 M. Use the dilution equation to determine the volume of the supplied phosphoric acid to be taken.
We have M1*V1 = M2*V2
where M1 = 1.00 M, M2 = 0.100 M and V2 = 100.00 mL; plug in and obtain
(1.00 M)*V1 = (0.100 M)*(100.00 mL)
=====> V1 = (0.100 M)*(100.00 mL)/(1.00 M) = 10.00 mL.
Next use the Henderson-Hasslebach equation to determine the ratio of the concentrations of the weak acid and the conjugate base as below.
pH = pKa + log [NaH2PO4.H2O]/[H3PO4]
=====> 2.50 = 2.12 + log [NaH2PO4.H2O]/[H3PO4]
=====> log [NaH2PO4.H2O]/[H3PO4] = 2.50 – 2.12 = 0.38
=====> [NaH2PO4.H2O]/[H3PO4] = antilog (0.38) = 2.3988
=====> [NaH2PO4.H2O] = 2.3988*[H3PO4]
Given [H3PO4] = 0.100 M in the buffer, we have,
[NaH2PO4.H2O] = 2.3988*[H3PO4] = 2.3988*(0.100 M) = 0.23988 M 0.240 M.
Since we have 100.00 mL of the buffer, hence, the mole(s) of NaH2PO4.H2O = (100.00 mL)*(1 L/1000 mL)*(0.240 M) = 0.024 mole.
Molar mass of NaH2PO4.H2O = (1*22.9897 + 2*1.008 + 1*30.9738 + 4*15.9994 + 2*2*1.008 + 2*1*15.9994) g/mol = 156.0079 g/mol.
Mass of NaH2PO4 taken = (0.024 mole)*(156.0079 g/mol) = 3.7442 g.
The buffer is prepared by pipetting 10.00 mL of the stock 1.00 M phosphoric acid in a 100.00 mL volumetric flask, adding 3.7442 g solid NaH2PO4.H2O and diluting upto the mark with DI water.
Weak Acid
Ka
pKa
Phosphoric acid, 1.00 M
7.52*10-3
2.12
Formic acid, 2.00 M
1.77*10-4
3.75
Sodium disulfate monohydrate, 3.00 M
1.20*10-2
1.92
Acetic acid, 5.00 M
1.75*10-5
4.76
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