Question

A buffer with a pH of 4.31 contains 0.31 M of sodium benzoate and 0.24 M of benzoic acid. What is the concentration of [ H 3 O ] in the solution after the addition of 0.060 mol HCl to a final volume of 1.3 L

Answer 1

Answer: The hydronium ion concentration in the solution is
`1.29* 10^(-4)M`

Step-by-step explanation:

To calculate the molarity, we use the equation:

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

Moles hydrochloric acid solution = 0.060 mol

Volume of solution = 1 L

Putting values in above equation, we get:

`\text{Molarity of HCl}=(0.060)/(1L)\\\\\text{Molarity of HCl}=0.060M`

The chemical reaction for aniline and HCl follows the equation:

`C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-`

Initial: 0.24 0.060 0.31

Final: 0.18 - 0.37

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

`pH=pK_a+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})`

`pH=pK_a+\log(([C_6H_5COO^-])/([C_6H_5COOH]))`

We are given:

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

`[C_6H_5COO^-]=0.18M`

`[C_6H_5COOH]=0.37M`

pH = ?

Putting values in equation 1, we get:

`pH=4.2+\log((0.18)/(0.37))\\\\pH=3.89`

To calculate the hydronium ion concentration in the solution, we use the equation:

`pH=-\log[H_3O^+]`

pH = 3.89

Putting values in above equation, we get:

`3.89=-\log[H_3O^+]`

`[H_3O^+]=10^(-3.89)=1.29* 10^(-4)M`

Hence, the hydronium ion concentration in the solution is
`1.29* 10^(-4)M`