Question

"" What is the pH of a solution that is prepared by dissolving 8.52 grams of lactic acid (formula weight = 90.08 grams/mol) and 7.93 grams of sodium lactate (formula weight = 112.06 grams/mole) in water and diluting to 500.00 mL? The Ka for lactic acid is 0.000137.

Answer 1

Answer: The pH of given solution is 3.74.

Step-by-step explanation:

The given data is as follows.

Mass of lactic acid = 8.52 g, Formula weight of lactic acid = 90.08 g/mol

So, number of moles of lactic acid will be calculated as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

=
`(8.52 g)/(90.08 g/mol)`

= 0.094 moles

Mass of sodium lactate = 7.93 g, Formula weight of sodium lactate = 112.06 g/mol

Hence, number of moles of sodium lactate is as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

=
`(7.93 g)/(112.06 g/mol)`

= 0.071 moles

As we know that relation between
`K_(a)` and
`pK_(a)` is as follows.

`pK_(a) = -log K_(a)`

= -log(0.000137)

= 3.86

Using Henderson equation, we will calculate the pH as follows.

pH =
`pK_(a) + log (\frac{\text{Conjugate base}}{\text{Acid}})`

pH =
`3.86 + log (\frac{\text{sodium lactate}}{\text{lactic acid}})`

=
`3.86 + log ((0.071)/(0.094))`

= 3.86 + log (0.755)

= 3.86 - 0.121

= 3.74

Therefore, we can conclude that pH of given solution is 3.74.