Question

A student performs a titration of 25.0 mL of 0.100 M lactic acid (HC3H5O3), using 0.050 M sodium hydroxide (NaOH). The Ka for lactic acid is 1.4 x 10-4. a) (10 points) What is the pH of the solution after the addition of 23.5 mL of sodium hydroxide solution

Answer 1

Answer: The pH of resulting solution is 3.815

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}`

• For lactic acid:

Molarity of lactic acid solution = 0.100 M

Volume of solution = 25.0 mL

Putting values in above equation, we get:

`0.100M=\frac{\text{Moles of lactic acid}* 1000}{25mL}\\\\\text{Moles of lactic acid}=(0.100* 25)/(1000)=0.0025mol`

• For NaOH:

Molarity of NaOH solution = 0.050 M

Volume of solution = 23.5 mL

Putting values in above equation, we get:

`0.050M=\frac{\text{Moles of NaOH}* 1000}{23.5mL}\\\\\text{Moles of NaOH}=(0.050* 23.5)/(1000)=0.0012mol`

The chemical reaction for NaOH and lactic acid follows the equation:

`HC_3H_5O_3+NaOH\rightarrow C_3H_5O_3Na+H_2O`

Initial: 0.0025 0.0012

Final: 0.0013 - 0.0012

Volume of solution = 25 + 23.5 = 48.5 mL = 0.0485 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

`pH=pK_a+\log(([salt])/([acid]))`

`pH=pK_a+\log(([C_3H_5O_3Na])/([HC_3H_5O_3]))`

We are given:

`pK_a` = negative logarithm of acid dissociation constant of lactic acid = 3.85

`[HC_3H_5O_3]=(0.0013)/(0.0485)`

`[C_3H_5O_3Na]=(0.0012)/(0.0485)`

pH = ?

Putting values in above equation, we get:

`pH=3.85+\log((0.0012/0.0485)/(0.0013/0.0485))\\\\pH=3.815`

Hence, the pH of resulting solution is 3.815