Question

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propionic acid (HC2H5CO2) with a 1.1000M solution of KOH. The pKa of proionic acid 4.89.

Calculate the pH of the acid solution after the chemist has added 41.04mL of the KOH solution to it.

Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added. Round your answer to decimal places.

Answer 1

Answer: The pH of acid solution is 4.58

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)}}` .....(1)

• For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

`1.1000M=\frac{\text{Moles of KOH}* 1000}{41.04}\\\\\text{Moles of KOH}=(1.1000* 41.04)/(1000)=0.04514mol`

• For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

`0.6100M=\frac{\text{Moles of propanoic acid}* 1000}{224.9}\\\\\text{Moles of propanoic acid}=(0.6100* 224.9)/(1000)=0.1372mol`

The chemical reaction for propanoic acid and KOH follows the equation:

`C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O`

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 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(\frac{[\text{salt}]}{[acid]})`

`pH=pK_a+\log(([C_2H_5COOK])/([C_2H_5COOH]))`

We are given:

`pK_a` = negative logarithm of acid dissociation constant of propanoic acid = 4.89

`[C_2H_5COOK]=(0.04514)/(0.26594)`

`[C_2H_5COOH]=(0.09206)/(0.26594)`

pH = ?

Putting values in above equation, we get:

`pH=4.89+\log(((0.04514/0.26594))/((0.09206/0.26594)))\\\\pH=4.58`

Hence, the pH of acid solution is 4.58