Question

What is the pH of 100 mL of a 50 mM solution of benzoic acid is added to 70 mL of a 50 mM NaOH solution?

Answer 1

Answer: The pH of the solution is 4.57

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 benzoic acid:

Molarity of benzoic acid = 50 mM = 0.05 M (Conversion factor: 1 M = 1000 mM)

Volume of solution = 100 mL

Putting values in above equation, we get:

`0.05M=\frac{\text{Moles of benzoic acid}* 1000}{100mL}\\\\\text{Moles of benozic acid}=((0.05* 100))/(1000)=0.005mol`

• For sodium hydroxide:

Molarity of sodium hydroxide = 50 mM = 0.05 M

Volume of solution = 70 mL

Putting values in above equation, we get:

`0.05M=\frac{\text{Moles of sodium hydroxide}* 1000}{50mL}\\\\\text{Moles of sodium hydroxide}=(0.05* 70)/(1000)=0.0035mol`

The chemical reaction for sodium hydroxide and benzoic acid follows the equation:

`C_6H_5COOH+NaOH\rightarrow C_6H_5COONa+H_2O`

Initial: 0.005 0.0035

Final: 0.0015 - 0.0035

Volume of solution = 100 + 70 = 170 mL = 0.170 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_6H_5COONa])/([C_6H_5COOH]))`

We are given:

`pK_a` = negative logarithm of acid dissociation constant of benzoic acid = 4.2

`[C_6H_5COONa]=(0.0035)/(0.170)`

`[C_6H_5COOH]=(0.0015)/(0.170)`

pH = ?

Putting values in above equation, we get:

`pH=4.2+\log((0.0035/0.170)/(0.0015/0.170))\\\\pH=4.57`

Hence, the pH of the solution is 4.57