Question

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.

Answer 1

Step-by-step explanation:

It is known that
`pK_(a)` value of acetic acid is 4.74. And, relation between pH and
`pK_(a)` is as follows.

pH = pK_{a} + log
`([CH_(3)COOH])/([CH_(3)COONa])`

= 4.74 + log
`(1.00)/(1.00)`

So, number of moles of NaOH = Volume × Molarity

= 71.0 ml × 0.760 M

= 0.05396 mol

Also, moles of
`CH_(3)COOH` = moles of
`CH_(3)COONa`

= 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.

`CH_(3)COONa + NaOH \rightarrow CH_(3)COOH`

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
`([CH_(3)COONa])/([CH_(3)COOH])`

= 4.74 + log
`(0.94604)/(1.05396)`

= 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.