Question

3. Formic acid, HCHO₂, has a K_a = 1.8 x 10^-4. We are asked to prepare a buffer solution having a pH of 3.40 from a 0.10 M HCHO₂ (formic acid solution) and 0.10 M NaCHO₂ (sodium formate; formate ion is the conjugate base). How many mL of the NaHCO₂ solution should be added to 20.0 mL of the 0.10 M HCHO₂ solution to make the buffer?

4. When 5 drops of 0.10 M NaOH were added to 20 mL of the buffer in problem 3, the pH went from 3.40 to 3.43. Write a net ionic equation to explain why the pH didn’t go up to about 10, as it would have if that amount of NaOH were added to distilled water or to about 20.0 mL of 0.00040 M HCl, which also would have a pH of 3.40.

Answer 1

3)

Ka = 1.8 x 10^-4

pKa = -log Ka

pKa = -log (1.8 x 10^-4)

pKa = 3.74

millmoles of HCHO2 = 20 x 0.1 = 2

millmoles of NaCHO2 = 0.1 V

pH = pKa + log [NaCHO2 / HCHO2]

3.40 = 3.74 + log (0.1 V / 2)

0.1 V / 2 = 0.457

V = 9.14 mL

volume of NaHCO2 needed = 9.1 mL

4)

net ionic equation :

HCHO2 + OH- -----------------------> CHO2-   + H2O

explantion : on addition of small amount of base the base react with acid forms conjugate base which is already part of buffer .so the pH doesnot go up