Question

1. The pH of a 50.0 mL solution of 0.500 M HSCN (the analyte) was measured and found to be 3.80. What is pKa for the acid based on this measurement?

2. Trace amounts of an Indicator HIn was added to the solution described above. The indicator has a pKa (pkIn) of 11.85. This indicator is known to be yellow in solution in the form HIn and blue in solution in the form In- . In a forward titration, a person with average eyesight will observe the colour change when there is a 1:10 ratio of In- : HIn in solution.

a. What is the ratio of In- : HIn in solution once the Indicator is mixed with the analyte?

b. What colour will the analyte solution appear once mixed with the indicator?

3. The solution described above was titrated with a 0.250 M NaOH solution.

a. What is the pH of the system at equivalence?

b. What is the ratio of In- : HIn in the system at equivalence?

c. What colour will the solution appear at equivalence?

4. Post-equivalence, the pH of the system will approach the pH of the titrant.

a. What is the pH the system will approach post equivalence?

b. What ratio of In- : HIn is expected in the solution at this pH?

c. What colour will the solution appear at this pH?

5. In the forward titration described above, at what pH would you expect the color transition to be observable for the first time?

thank you for the help :)

Answer 1

For weak acid

pH = 3.80

[H+] = 0.000158 M

Ka =( 0.000158)²/(0.50-0.000158)

= 5.02×10^-8