Question

An unknown acid was titrated two times with 0.10 M NaOH. The first titration was done using an indicator to determine the equivalence point volume. A second titration was carried out where the pH was determined throughout the entire titration. The data is given below.
Titration 1: mL unknown acid = 25.0, mL of NaOH needed to reach equivalence point = 30.00
Titration 2: mL unknown acid = 25.0
mL 0.1M pH
NaOH Added
0.0 3.72
5.0 5.82
10.0 6.22
15.0 6.52
20.0 6.82
25.0 7.22
30.0 9.63
35.0 11.92
a. What is the concentration of the unknown acid?
b. What is the Ka of the unknown acid?
c. Which of the indicators given below would be a good choice to use in the titration of this acid with the NaOH?

Answer 1

Answer:

hello your question is incomplete attached below is the complete question

a) 0.12 M

b) Ka = 3.0 * 10^-7

c) Alizarin Yellow R

Explanation:

A) Determine the concentration of the unknown acid

The PH of the unknown acid before addition of NaOH is ; 3.72 ( weak acid )

First determine the moles of of NaOH

= molarity * volume

= 0.10 M * 30.0 ML = 0.0030 mol

at equivalence point

moles of NaOH = moles of unknown acid = 0.0030 mol

volume of unknown acid = 25.0 mL

Next calculate the concentration of the Acid ( HA )

= moles / volume

= 0.0030 mol / 25.0 mL

= 0.12 M

b) Determine the Ka of the unknown acid

attached below is the detailed solution

Ka = 3.0 * 10^-7

c) The indicator that would be a good choice to use in the titration of this acid with NaOH is ; Alizarin Yellow R . this is because the titration is between a strong base and a weak acid.