Question

(b) what is the ph of a solution containing 0.03m benzoic acid (pka 3.19) and 0.02m nabenzoate?

Answer 1

To calculate the pH of the solution containing 0.03 M benzoic acid and 0.02 M sodium benzoate, the Henderson-Hasselbalch equation is used with a pKa value of 3.19, resulting in a pH of approximately 3.0139.

Step-by-step explanation:

To calculate the pH of a solution containing 0.03 M benzoic acid and 0.02 M sodium benzoate, we can use the Henderson-Hasselbalch equation, which is given as:

pH = pKa + log([A-]/[HA])

Where [A-] represents the concentration of the conjugate base (sodium benzoate) and [HA] represents the concentration of the acid (benzoic acid). Given that the pKa of benzoic acid is 3.19, we can substitute the values into the equation:

pH = 3.19 + log(0.02/0.03)

Calculating the log(0.02/0.03) gives us a value of approximately -0.1761. When added to the pKa value of 3.19:

pH = 3.19 - 0.1761

Therefore, the pH of the solution is approximately 3.0139.

Answer 2

Answer is: pH of solution is 3,02.
c(benzoic acid) = 0,03 mol/dm³.
c(sodium benzoate) = 0,02 mol/dm³.
pKa(benzoik acid) = 3,19.
pH = ?
Henderson–Hasselbalch equation for buffers: pH = pKa + log(cs/ck)
pH = 3,19 + log(0,02/0,03)
pH = 3,19 - 0,176
pH = 3,02.
cs - concentration of acid.