Question

In the laboratory, a general chemistry student measured the pH of a 0.329 M aqueous solution of benzoic acid, C6H5COOH to be 2.327. Use the information she obtained to determine the Ka for this acid. Ka(experiment) =

Answer 1

Answer:

The dissociation constant for the acid ( experimental ) is 1.45 lit/mol

Explanation:

The value of dissociation constant can be calculated as,

  $K_{a}$ = C × ∝²

Where, C = concentration of the solution = 0.329M

          ∝ = Degree of dissociation

again , Degree of dissociation can be obtained form :

                      $p_{H}$ = C × ∝

                         ∝ = $\frac{p_{H} }{C}$

                        ∝ = $\frac{2.327}{0.329}$ = 7.072

So, now $K_{a}$ = C × ∝²

                     = 0.329 ×( 7.072)²

                     = 1.45 lit/ mol

Answer 2

Answer:

The Ka = 6.74 * 10^-5

Explanation:

Step 1: Data given

Concentration of benzoic acid = 0.329 M

pH = 2.327

Step 2: Calculate the Ka

pH = -log (√([HA]*Ka))

2.327 = -log (√(0.329*Ka))

10 ^ - 2.327 = √(0.329*Ka))

0.0047098 = √(0.329*Ka))

2.218 * 10^-5 = 0.329 * Ka

Ka = 6.74 * 10^-5

The Ka = 6.74 * 10^-5