Question

c) A student titrated 50.0mL of a 0.10M solution of a certain weak acid with NaOH(aq) . The results are given in the graph above. (i) What is the approximate pKa of the acid?

Answer 1

3.2

Step-by-step explanation:

The missing graph is attached in the image below:

Initially, p.H from the graph is approximately 2.1

Therefore; the hydrogen ion concentration is

`[H^+] = 10^(-2.1)`

`[H^+] = ({k_a * 0.10})^{(1)/(2)}`

`{k_a} = 6.3 * 10^( -4)`

pKa =
`-log ( 6.3 * 10^(-4))`

pKa = 3.2