Abuffer contains 0.19 mol of propionic acid (c2h5cooh) and 0.26 mol of sodium propionate (c2h5coona) in 1.20 l. you may want to reference (pages 721 - 729) section 17.2 while completing this problem. part a what is the ph of this buffer? express the ph to two decimal places. php h = nothing request answer part b what is the ph of the buffer after the addition of 0.02 mol of naoh? express the ph to two decimal places. php h = nothing request answer part c what is the ph of the buffer after the addition of 0.02 mol of hi? express the ph to two decimal places.

Explanation:

It is known that of propionic acid = 4.87

And, initial concentration of  propionic acid =

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH =

                       =

                        = 4.87 + log (1.36)

                        = 5.00

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid =

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH =

                       =

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid =

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH =

                       =

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

c. electrical charges of ionic bonds.