Question

Lactic acid C_3H_6O_3 is found in sour milk where it is produced by the action of lactobacilli in lactose or the sugar in milk. A 0.025 M solution of lactic acid has a pH of 2.75. Calculate the equilibrium constant for this acid Had you not been given the pH of the acid and you have to measure it yourself, how would the method of this experiment be applied to the determination of Ka? Would you expect an improvement in the accuracy of your result with the application of the method of this experiment? Explain why or why not.

Answer 1

Step-by-step explanation:

• Let lactic acid be represent as HA. And, pH is given as 2.75.

Hence, concentration of hydrogen ions will be calculated as follows.

pH =
`-log[H^(+)]`

`[H^(+)] = 10^(-2.75)`

=
`1.78 * 10^(-3)` M

= 0.00178 M

The reaction equation will be as follows.

`HA \rightarrow H^(+) + A^(-)`

Initial: 0.025 0 0

Change: -0.00178 0.00178 0.00178

Equilibrium: 0.0232 0.00178 0.00178

Hence, calculate the value of
`K_(a)` as follows.

`K_(a) = ([H^(+)][A^(-)])/([HA])`

=
`(0.00178 * 0.00178)/(0.0232)`

=
`1.37 * 10^(-4)`

Therefore, value of the equilibrium constant is
`1.37 * 10^(-4)` .

• Another method that we can use to determine the value of
  `K_(a)` of any weak acid is when weak acid is titrating with a strong base then,

number of moles of acid = moles of salt formed by (acid + base)

These moles of salt are actually known as half equivalence point. So, at this point we need we need to measure the pH of solution with the help of a device called pH meter.

As we known that pH =
`pK_(a)`

and,
`K_(a) = 10^{-pK_(a)}`