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.

idk i’m struggling too

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.