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) =

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

answer: you could do some invislibe ink.

explanation: i will tell you how to if u answer my question.

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying