Question

How would you prepare 50.0 mL of a NaChi0Na CHO. buffer with pH-635 (give spec each component)? Assume that you want the total buffer concentration (Le, [weak acid]+ [conjugate ific masses of base]) 6、 a. to be 0.100 M. Review Example 16.4 on p. 72 in the Chang textbook. Make sure to select the sppropriate K- value for this buffer system. Show your work below. Maleic Acid: p-1.90 p-6.30 pk.--log K) [weak acid]+[conjugate base]- 0.100 M pH- pKa + MW (NaCH04) - 138.05 g/mol For this buffer: log (wenk acid (conjugate base] MW (Na,CaH04)- 160.04 g/mol Mass of NaC4HO4 needed- Mass of Na C4H O4 needed- b. If 0.0010 moles of NaOH are added to the buffer in part a, what would happen to the pH (increase a little, decrease a little, or stay the same)? Explain your answer (no calculation is required here but drawing a picture or writing chemical reactions might be helpful). Calculate the pH of the solution in part b. (Review Example 16.3 (pp. 719-720) of Chang) c.

Answer 1

Q6a) let concentration of Na2C4H2O4 = s mol/L and that of NaC4H3O4 = (0.100 - s)mol/L

pH = pK_a2 + log([Na2C4H2O4]/[NaC4H3O4])

6.35 = 6.30 + log (s/0.100-s)

Log(s/(0.100-s))= 0.05

s/(0.100-s) = 1.122

s = 0.0529mol/L

[NaC4H3O4] = 0.100 - 0.0529 = 0.0471mol/L

Mole of Na2C4H2O4 = 0.0529mol/L × 50.0×10^-3L

= 2.645×10^-3mol

Mass of Na2C4H2O4 = 2.645×10^-3mol × 160.04g/mol

= 0.4233g (answer)

Mass of NaC4H3O4 = 0.0471mol/L × 50.0×10^-3L × 138.05g/mol = 0.3251g (answer)

Q6b)

On adding NaOH , it will react with acid.

NaOH + NaC4H3O4 $\rightarrow$ Na₂C4H2O4 + H₂O.

pH = pK_a2 + log([Na2C4H2O4]/[NaC4H3O4])

Concentration of Na2C4H2O4 increases so pH will also rise.

pH of solution will increases a little. (answer)

Q6c)

Moles of Na2C4H2O4 = 0.002645 + 0.0010 = 0.003645

Moles of NaC4H3O4 = 0.002355 + 0.0010 = 0.001355

pH = 6.30 + log (0.003645/0.001355)

= 6.73