Calculate the change in ph when 71.0 ml of a 0.760 m solution of naoh is added to 1.00 l of a solution that is 1.0o m in sodium acetate and 1.00 m in acetic acid.

Explanation:

It is known that value of acetic acid is 4.74. And, relation between pH and is as follows.

                    pH = pK_{a} + log

                          = 4.74 + log

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of   = moles of

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log

               = 4.74 +  log

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

you did not give any options thus this is the best answer i can give to you with this little information you have given

> they are both natural and not man made

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