Question

Sodium benzoate (C6H5COONa), the sodium salt of the weak acid benzoic acid, is used as a food preservative. A solution is prepared by dissolving 0.100 mol of sodium benzoate in enough pure water to produce 1.00 L of solution. If the pKa for benzoic acid is 4.20, calculate the pH of the sodium benzoate solution.

Answer 1

8.6

Step-by-step explanation:

Using the expression :

`K_a* K_b=K_w`

Where,
`K_w` is the dissociation constant of water.

At
`25\ ^0C`,
`K_w=10^(-14)`

Thus, for benzoic acid , pKa = 4.20

Thus,
`K_a=10^(-4.20)=6.31* 10^(-5)`

`K_b` for Sodium benzoate can be calculated as:

`K_a* K_b=K_w`

`6.31* 10^(-5)* K_b=10^(-14)`

`K_b=1.58* 10^(-10)`

The benzoate ion will dissociate as:-

`C_6H_5COO^-_((aq)) + H_2O_((l))\rightleftharpoons C_6H_5COOH_((aq)) + OH^-_((aq))`

`K_b` expression is:-

`K_(b)=\frac {\left [ C_6H_5COOH^(+) \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}`

Given that:-

Moles = 0.100 moles

Volume = 1.00 L

Thus, Concentration = 0.100/ 1.00 M = 0.1 M

Considering the ICE table in the image below.

So,

`1.58* 10^(-10)=(x^2)/(0.1-x)`

`1.58\left(0.1-x\right)=10^(10)x^2`

Solving for x, we get that,
`x=3.97* 10^(-6)`

Thus,
`[OH^-]=3.97* 10^(-6)`

Also,

`pOH=-log[OH^-]=-log(3.97* 10^(-6))=5.4`

Also, pH + pOH = 14

So, pH = 14 - 5.4 = 8.6