Question

A buffer 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.

Answer 1

Explanation:

It is known that $pK_{a}$ of propionic acid = 4.87

And, initial concentration of  propionic acid = $\frac{0.19}{1.20}$

                                                                       = 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 = $pK_{a} + log(\frac{[salt]}{[acid]})$

                       = $4.87 + log \frac{0.216}{0.158}$

                        = 4.87 + log (1.36)

                        = 5.00

• Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = $\frac{0.17}{1.20 L}$

                                                                 = 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.

                        $\frac{0.28 mol}{1.20 L}$

                           = 0.23 M

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

             pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

                       = $4.87 + log \frac{0.23}{0.14}$

                        = 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.

• Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = $\frac{0.21}{1.20 L}$

                                                                 = 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.

                        $\frac{0.24 mol}{1.20 L}$

                           = 0.2 M

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

             pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

                       = $4.87 + log \frac{0.2}{0.175}$

                        = 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.